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Very-long-baseline interferometry - Wikipedia
Comparing widely separated telescope wavefronts
VLBI was used to create the first image of a black hole, imaged by the Event Horizon Telescope and published in April 2019.[1]
Data received at each antenna in the array include arrival times from a local atomic clock, such as a hydrogen maser. At a later time, the data are correlated with data from other antennas that recorded the same radio signal, to produce the resulting image. The resolution achievable using interferometry is proportional to the observing frequency. The VLBI technique enables the distance between telescopes to be much greater than that possible with conventional interferometry, which requires antennas to be physically connected by coaxial cable, waveguide, optical fiber, or other type of transmission line. The greater telescope separations are possible in VLBI due to the development of the closure phase imaging technique by Roger Jennison in the 1950s, allowing VLBI to produce images with superior resolution.[2]
VLBI is best known for imaging distant cosmic radio sources, spacecraft tracking, and for applications in astrometry. However, since the VLBI technique measures the time differences between the arrival of radio waves at separate antennas, it can also be used "in reverse" to perform earth rotation studies, map movements of tectonic plates very precisely (within millimetres), and perform other types of geodesy. Using VLBI in this manner requires large numbers of time difference measurements from distant sources (such as quasars) observed with a global network of antennas over a period of time.
6 International VLBI Service for Geodesy and Astrometry
In VLBI, the digitized antenna data are usually recorded at each of the telescopes (in the past this was done on large magnetic tapes, but nowadays it is usually done on large arrays of computer disk drives). The antenna signal is sampled with an extremely precise and stable atomic clock (usually a hydrogen maser) that is additionally locked onto a GPS time standard. Alongside the astronomical data samples, the output of this clock is recorded. The recorded media are then transported to a central location. More recent[when?] experiments have been conducted with "electronic" VLBI (e-VLBI) where the data are sent by fibre-optics (e.g., 10 Gbit/s fiber-optic paths in the European GEANT2 research network) and not recorded at the telescopes, speeding up and simplifying the observing process significantly. Even though the data rates are very high, the data can be sent over normal Internet connections taking advantage of the fact that many of the international high speed networks have significant spare capacity at present.
At the location of the correlator, the data are played back. The timing of the playback is adjusted according to the atomic clock signals, and the estimated times of arrival of the radio signal at each of the telescopes. A range of playback timings over a range of nanoseconds are usually tested until the correct timing is found.
Each antenna will be a different distance from the radio source, and as with the short baseline radio interferometer the delays incurred by the extra distance to one antenna must be added artificially to the signals received at each of the other antennas. The approximate delay required can be calculated from the geometry of the problem. The tape playback is synchronized using the recorded signals from the atomic clocks as time references, as shown in the drawing on the right. If the position of the antennas is not known to sufficient accuracy or atmospheric effects are significant, fine adjustments to the delays must be made until interference fringes are detected. If the signal from antenna A is taken as the reference, inaccuracies in the delay will lead to errors
{\displaystyle \epsilon _{B}}
{\displaystyle \epsilon _{C}}
in the phases of the signals from tapes B and C respectively (see drawing on right). As a result of these errors the phase of the complex visibility cannot be measured with a very-long-baseline interferometer.
Temperature variations at VLBI sites can deform the structure of the antennas and affect the baseline measurements.[3][4] Neglecting atmospheric pressure and hydrological loading corrections at the observation level can also contaminate the VLBI measurements by introducing annual and seasonal signals, like in the Global Navigation Satellite System time series.[4]
Roger Clifton Jennison developed a novel technique for obtaining information about visibility phases when delay errors are present, using an observable called the closure phase. Although his initial laboratory measurements of closure phase had been done at optical wavelengths, he foresaw greater potential for his technique in radio interferometry. In 1958 he demonstrated its effectiveness with a radio interferometer, but it only became widely used for long-baseline radio interferometry in 1974. At least three antennas are required. This method was used for the first VLBI measurements, and a modified form of this approach ("Self-Calibration") is still used today.
Imaging the surfaces of nearby stars at radio wavelengths (see also interferometry) – similar techniques have also been used to make infrared and optical images of stellar surfaces.
Definition of the celestial reference frame.[5][6]
Motion of the Earth's tectonic plates.
Earth's orientation parameters and fluctuations in the length of day.[7]
Maintenance of the terrestrial reference frame.
Measurement of gravitational forces of the Sun and Moon on the Earth and the deep structure of the Earth.
Measurement of the fundamental speed of gravity.
The tracking of the Huygens probe as it passed through Titan's atmosphere, allowing wind velocity measurements.[8]
First imaging of a supermassive black hole.[1][9]
VLBI arrays[edit]
There are several VLBI arrays located in Europe, Canada, the United States, Chile, Russia, China, South Korea, Japan, Mexico, Australia and Thailand. The most sensitive VLBI array in the world is the European VLBI Network (EVN). This is a part-time array that brings together the largest European radiotelescopes and some others outside of Europe for typically weeklong sessions, with the data being processed at the Joint Institute for VLBI in Europe (JIVE). The Very Long Baseline Array (VLBA), which uses ten dedicated, 25-meter telescopes spanning 5351 miles across the United States, is the largest VLBI array that operates all year round as both an astronomical and geodesy instrument.[10] The combination of the EVN and VLBA is known as Global VLBI. When one or both of these arrays are combined with space-based VLBI antennas such as HALCA or Spektr-R, the resolution obtained is higher than any other astronomical instrument, capable of imaging the sky with a level of detail measured in microarcseconds. VLBI generally benefits from the longer baselines afforded by international collaboration, with a notable early example in 1976, when radio telescopes in the United States, USSR and Australia were linked to observe hydroxyl-maser sources.[11] This technique is currently being used by the Event Horizon Telescope, whose goal is to observe the supermassive black holes at the centers of the Milky Way Galaxy and Messier 87.[1][12][13]
e-VLBI[edit]
VLBI has traditionally operated by recording the signal at each telescope on magnetic tapes or disks, and shipping those to the correlation center for replay. Recently,[when?] it has become possible to connect VLBI radio telescopes in close to real-time, while still employing the local time references of the VLBI technique, in a technique known as e-VLBI. In Europe, six radio telescopes of the European VLBI Network (EVN) are now connected with Gigabit per second links via their National Research Networks and the Pan-European research network GEANT2, and the first astronomical experiments using this new technique were successfully conducted in 2011.[14]
Space VLBI[edit]
In the quest for even greater angular resolution, dedicated VLBI satellites have been placed in Earth orbit to provide greatly extended baselines. Experiments incorporating such space-borne array elements are termed Space Very Long Baseline Interferometry (SVLBI). The first SVLBI experiment was carried out on Salyut-6 orbital station with KRT-10, a 10-meter radio telescope, which was launched in July 1978.[citation needed]
The first dedicated SVLBI satellite was HALCA, an 8-meter radio telescope, which was launched in February 1997 and made observations until October 2003. Due to the small size of the dish, only very strong radio sources could be observed with SVLBI arrays incorporating it.
Another SVLBI satellite, a 10-meter radio telescope Spektr-R, was launched in July 2011 and made observations until January 2019. It was placed into a highly elliptical orbit, ranging from a perigee of 10,652 km to an apogee of 338,541 km, making RadioAstron, the SVLBI program incorporating the satellite and ground arrays, the biggest radio interferometer to date. The resolution of the system reached 8 microarcseconds.
International VLBI Service for Geodesy and Astrometry[edit]
The International VLBI Service for Geodesy and Astrometry (IVS) is an international collaboration whose purpose is to use the observation of astronomical radio sources using (VLBI) to precisely determine earth orientation parameters (EOP) and celestial reference frames (CRF) and terrestrial reference frames (TRF).[15] IVS is a service operating under the International Astronomical Union (IAU) and the International Association of Geodesy (IAG).[16]
^ a b c The Event Horizon Telescope Collaboration (April 10, 2019). "First M87 Event Horizon Telescope Results. I. The Shadow of the Supermassive Black Hole". The Astrophysical Journal Letters. 875 (1): L1. arXiv:1906.11238. Bibcode:2019ApJ...875L...1E. doi:10.3847/2041-8213/ab0ec7.
^ R. C. Jennison (1958). "A Phase Sensitive Interferometer Technique for the Measurement of the Fourier Transforms of Spatial Brightness Distributions of Small Angular Extent". Monthly Notices of the Royal Astronomical Society. 119 (3): 276–284. Bibcode:1958MNRAS.118..276J. doi:10.1093/mnras/118.3.276.
^ Wresnik, J.; Haas, R.; Boehm, J.; Schuh, H. (2007). "Modeling thermal deformation of VLBI antennas with a new temperature model". Journal of Geodesy. 81 (6–8): 423–431. Bibcode:2007JGeod..81..423W. doi:10.1007/s00190-006-0120-2. S2CID 120880995.
^ a b Ghaderpour, E. (2020). "Least-squares wavelet and cross-wavelet analyses of VLBI baseline length and temperature time series: Fortaleza-Hartrao-Westford-Wettzell". Publications of the Astronomical Society of the Pacific. 133: 1019. doi:10.1088/1538-3873/abcc4e. S2CID 234445743.
^ "The ICRF". IERS ICRS Center. Paris Observatory. Retrieved 25 December 2018.
^ "International Celestial Reference System (ICRS)". United States Naval Observatory. Retrieved 25 December 2018.
^ Urban, Sean E.; Seidelmann, P. Kenneth, eds. (2013). Explanatory Supplement to the Astronomical Almanac, 3rd Edition. Mill Valley, California: University Science Books. pp. 176–7. ISBN 978-1-891389-85-6.
^ Clery, Daniel (April 10, 2019). "For the first time, you can see what a black hole looks like". Science. AAAS. Retrieved April 10, 2019.
^ "Very Long Baseline Array (VLBA)". National Radio Astronomy Observatory. Archived from the original on June 11, 2012. Retrieved May 30, 2012.
^ First Global Radio Telescope, Sov. Astron., Oct 1976
^ Webb, Jonathan (8 January 2016). "Event horizon snapshot due in 2017". bbc.com. BBC News. Retrieved 2017-10-22.
^ "Astronomers Demonstrate a Global Internet Telescope". Retrieved 2011-05-06.
^ Nothnagel, A.; Artz, T.; Behrend, D.; Malkin, Z. (8 September 2016). "International VLBI Service for Geodesy and Astrometry". Journal of Geodesy. 91 (7): 711–721. Bibcode:2017JGeod..91..711N. doi:10.1007/s00190-016-0950-5. S2CID 123256580.
^ Schuh, H.; Behrend, D. (October 2012). "VLBI: A fascinating technique for geodesy and astrometry". Journal of Geodynamics. 61: 68–80. Bibcode:2012JGeo...61...68S. doi:10.1016/j.jog.2012.07.007. hdl:2060/20140005985.
Wikimedia Commons has media related to Very-long-baseline interferometry.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Very-long-baseline_interferometry&oldid=1087576581"
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Headspace Moisture Analysis in Medical Device Packages | J. Med. Devices | ASME Digital Collection
Bruce Lindner,
, Oakdale, USA
Jonas Weissenrieder,
Lindner, B., Curtis, M., Weissenrieder, J., and Schwab, M. (June 11, 2008). "Headspace Moisture Analysis in Medical Device Packages." ASME. J. Med. Devices. June 2008; 2(2): 027504. https://doi.org/10.1115/1.2927393
Moisture levels in medical device packages influence a variety of crucial device properties, e.g. mechanical properties, corrosion and leach rates, drug potency, and ultimately shelf life. This is especially true for drug releasing and biodegradable device materials. It is therefore important to establish a high degree of control and accuracy of the humidity levels at all relevant stages in the production process as well as in the final package. In the current study we demonstrate a newly developed method for accurate headspace moisture trace level analysis in medical device packages using extractive gas phase Fourier transform infrared (FTIR) spectroscopy. Volumetric aliquots were extracted, using a specially designed extraction assembly, from the headspace of medical device relevant packages. The headspace water concentration was analyzed using a validated gas phase FTIR system1. Water bands in the spectral region
1600–2200cm
1 were chosen for the quantitative analysis. Sample spectra were compared, with a least square fit procedure, to water reference spectra at known concentration. Accurate quantification was demonstrated for headspace water vapor concentrations less than
100ppm
. This is considerably lower than feasible with conventional package headspace moisture analysis techniques. The results of this study demonstrate the benefits of using extractive gas phase FTIR for low level moisture analysis of small headspace volumes.
biodegradable materials, biomedical equipment, chemical variables measurement, corrosion, drugs, Fourier transform spectroscopy, humidity measurement, infrared spectroscopy, least squares approximations, packaging, spectrochemical analysis
Corrosion, Drugs, Medical devices, Spectroscopy, Fourier transform infrared spectroscopy, Water, Manufacturing, Spectra (Spectroscopy), Biodegradable materials, Biodegradation, Biomedical equipment, Fourier transform spectroscopy, Fourier transforms, Infrared spectroscopy, Least squares approximations, Mechanical properties, Packaging, Water vapor
Hermeticity Assessment of MEMS Micro Packages: Leak Rate Measurements Based on Infrared Spectroscopy
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{\displaystyle n}
{\displaystyle {\frac {1}{n}}>{\frac {1}{n+1}}+{\frac {1}{{(n+2)}^{2}}}+{\frac {3}{{(n+3)}^{3}}}+{\frac {4^{2}}{{(n+4)}^{4}}}+{\frac {5^{3}}{{(n+5)}^{5}}}+\dots }
approximately whe{\displaystyle n}
{\displaystyle {\frac {1}{1001}}+{\frac {1}{1002^{2}}}+{\frac {3}{1003^{3}}}+{\frac {4^{2}}{1004^{4}}}+{\frac {5^{3}}{1005^{5}}}+\dots <{\frac {1}{1000}}}
{\displaystyle 10^{-440}}
Quotes about RamanujanEdit
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Stability factor μ of two-port network - MATLAB stabilitymu - MathWorks España
stabilitymu
Stability Factor of Two-Port Network
Examine Stability Factor Using S-Parameter Object
Stability factor μ of two-port network
[mu,muprime] = stabilitymu(s_params)
[mu,muprime] = stabilitymu(hs)
[mu,muprime] = stabilitymu(s_params) calculates and returns the stability factor, μ, and μ' for the two-port S-parameters
[mu,muprime] = stabilitymu(hs) calculates and returns the stability factor for the two-port network represented by the S-parameter object hs.
Calculate the stability factor of network data from a file.
s_params = S.Parameters;
[mu,muprime] = stabilitymu(s_params);
stability_index = (mu>1)|(muprime>1);
is_stable = all(stability_index)
is_stable = logical
List frequencies with unstable S-parameters.
freq_unstable = freq(~stability_index)
freq_unstable =
Create a sparameters object from the specified file.
s_params = sparameters('passive.s2p');
Calculate the stability factor using stabilitymu function.
Check the stability criteria.
freq = s_params.Frequencies;
mu — Minimum distance between center of unit Smith chart and unstable region in load plane
Minimum distance between the center of the unit Smith chart and the unstable region in the load plane, returned as vector equal to the number of frequency or data points.
muprime — Minimum distance between center of unit Smith chart and unstable region in source plane
Minimum distance between the center of the unit Smith chart and the unstable region in the source plane, returned as a vector equal to the number of frequency or data points.
stabilitymu calculates the stability factors using the equations
\begin{array}{c}\mathrm{μ}=\frac{1â{|{S}_{11}|}^{2}}{|{S}_{22}â{S}_{11}^{*}\mathrm{Î}|+|{S}_{21}{S}_{12}|}\\ \mathrm{μ}â²=\frac{1â{|{S}_{22}|}^{2}}{|{S}_{11}â{S}_{22}^{*}\mathrm{Î}|+|{S}_{21}{S}_{12}|}\end{array}
S11, S12, S21, and S22 are S-parameters, from the input argument s_params.
Δ is a vector whose members are the determinants of the M 2-port S-parameter matrices:
\mathrm{Î}={S}_{11}{S}_{22}â{S}_{12}{S}_{21}
S* is the complex conjugate of the corresponding S-parameter.
The function performs these calculations element-wise for each of the M S-parameter matrices in s_params.
[1] Edwards, M.L., and J.H. Sinsky. “A New Criterion for Linear 2-Port Stability Using a Single Geometrically Derived Parameter.†IEEE Transactions on Microwave Theory and Techniques 40, no. 12 (December 1992): 2303–11. https://doi.org/10.1109/22.179894.
stabilityk
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Liquid And Gaseous States, Popular Questions: Jee Online course CHEMISTRY, Chemistry - Meritnation
Learner asked a question
. 10g each of CH4 and O2 are kept in cylinders of same volume under same temperatures, give the pressure ratio of two gases.
Kindly explain the above question and its answer
Among H2, He, CH4 and O2, the gas which can be easily liquefied is
D)O2
compare average molar kinetic energies of He and Ar at 300K
1)K(He)=K(Ar)
2)K(He)>K(Ar)
3)K(He)=1/2K(Ar)
if i collect water sample from different deapts of sea will it contain same dissolved O2 give reson as well ?
what effcts solubility of gas in water
Please explain in detail.
A 28. A capillary of radius r is lowered into a wetting agent with surface tension a and density d.
(i) The height h0 to witch the liquid will rise in the capillary is -
(A)
\left[\frac{2a}{dgr}\right]
(B)
\left[\frac{4a}{dgr}\right]
(C)
\left[\frac{8a}{dgr}\right]
(D)
\left[\frac{a}{8dgr}\right]
(ii) The work done by surface tension is -
\left[\frac{2{\mathrm{\pi a}}^{2}}{dg}\right]
(B)
\left[\frac{4{\mathrm{\pi a}}^{2}}{dg}\right] \left(C\right) \left[\frac{8{\mathrm{\pi a}}^{2}}{dg}\right]
\left[\frac{12{\mathrm{\pi a}}^{2}}{dg}\right]
(iii) The potential energy acquired by the liquid in the capillary is equal to -
(A)
\left[\frac{4{\mathrm{\pi a}}^{2}}{dg}\right]
\left[\frac{6{\mathrm{\pi a}}^{2}}{dg}\right]
\left(C\right) \left[\frac{2{\mathrm{\pi a}}^{2}}{dg}\right] \left(\mathrm{D}\right) \left[\frac{7{\mathrm{\pi a}}^{2}}{dg}\right]
Why is option (b) the answer & not (d)?
Please explain the solution in detail.
Q5. An orange solid (X) on heating, gives a colourless gas (Y) and a only green residue (Z). Gas (Y) on treatment Mg, produces a white solid substance.....
(A) Mg3N2 (B) MgO (C) Mg2O3 (D) MgCl2
Please explain question 18 in detail.
In step 2, why does electrophilic substitution take place in positions 1 & 2?
7) What is the effective area of charcoal occupied per ammonia molecule.
\left(A\right)\frac{1}{2}\sqrt{3}×{10}^{-20} {m}^{2}\phantom{\rule{0ex}{0ex}}\left(B\right) 4\sqrt{3}×{10}^{-20} {m}^{2}\phantom{\rule{0ex}{0ex}}\left(C\right) 2\sqrt{3}×{10}^{-20} {m}^{2}\phantom{\rule{0ex}{0ex}}\left(D\right) 8\sqrt{3}×{10}^{-20} {m}^{2}\phantom{\rule{0ex}{0ex}}
Please explain question (11) in detail. (Assertion-Reason) type-
What does radioactivity mean in this question numerically?
Please explain why the 2nd reaction is not possible?
when gas becomes more compressible/liquefable at more or less a,b (vanderwalls constant) also at z>/<1
Snehasree Acharya asked a question
50 ml of avessel contains 30g of water.the vessel is saturated with nitrogen gas. the lose in water concentretion is 1.12g. calculate the aqueos tension.
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Finding the number of solutions to equations | StudyPug
m = \frac{y_2-y_1}{x_2- x_1}
m = \frac{y_2-y_1}{x_2- x_1}
Depending on whether and how the linear equations in a system touch each other, there will be different number of solutions to the system. There can be one solution, no solution and even infinite solution.
m = \frac{y_2-y_1}{x_2- x_1}
, Slope intercept form: y = mx + b, Parallel line equation
Related Concepts: System of linear-quadratic equations, System of quadratic-quadratic equations, Graphing systems of linear inequalities, Graphing systems of quadratic inequalities
\bullet
The solutions to a system of equations are the points of intersection of the graphs.
\bullet
For a system consisting of two linear equations:
State whether each of the following systems have ONE, NONE, or INFINITE solutions
ii) 6x + 2y = 10
iii) x - y = 3
Find a value for c that will give the following system:
3y + 2cx = 6
i) one solution
ii) no solutions
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Solving a system of linear equations with graphs | StudyPug
m = \frac{y_2-y_1}{x_2- x_1}
Graphing linear functions using various forms
m = \frac{y_2-y_1}{x_2- x_1}
When studying linear algebra two topics are of utmost importance: Notation of matrices and vector fields. This lesson will focus on concepts which are the base to understand vector field mathematics, since you need to know how functions are graphed, what type of variables are involved on them and make sense of the meaning behind their visual representations.
Given that our lesson for today will focus on graphing equations, there is a basic concept you must understand: ordered pairs. An ordered pair is a set of two values usually written inside of a parenthesis and separated by a coma. The function of an ordered pair is to describe the position of a point in a graph providing the abscissa and the ordinate coordinate points.
In other words, to graph an equation and thus using graphing as a method to solve a system of linear equations, it is necessary to obtain the values of the abscissa and ordinate coordinates equivalent to the values of the variables x and y from the equations. Such array of values will then be written as an array of ordered pairs (
x
y
) which can be graphed.
On the first section of this lesson let us first look at the different algebraic methods that can be used to solve a system of equations, then, the second and third sections will focus on graphing.
There are three different methods we use in algebra to solve a system of equations:
Solving systems of linear equations by elimination is one of the simplest methods. Its basic principle relies on adding or subtracting one equation from the other and thus quickly eliminate one of the two variables from the system so the variable that is left can be solved for automatically. Although this is probably the simplest method to use, is not always the most practical, and so it is left for systems with equations which contain the same coefficient on one of their variables. Examples of that can be seen in the lessons linked here.
The second method consists on solving systems of linear equations by substitution. Substitution means that we use one of the equations in the system to solve for one of its variables. Once you have an expression equal to the selected variable, you substitute this expression on the other equation in place of the variable so in that way, you end up with an equation in terms of the other variable which can be solved quickly to its final value.
The substitution method is considered to be the most difficult out of the three options, but as long as you follow the rule of: any operation done on one side of the equation, must be done to the other side this is the most used method by mathematicians. Substitution actually provides a basis for most of the algebraic methods to operate in any kind of mathematical function.
Our third method refers to solving systems of linear equations by graphing and is the topic we are covering in full today. For that, we start by explaining how an equation is graphed, to then finish our lesson with the method to solve systems of equations.
In order for us to graph a linear equation we need to solve for the dependent variable.
For every equation with two variables, there will always be an independent variable and a dependent variable, as a standard in mathematics, our dependent variable is the one we call
y
This means that the values
y
takes depend on the value of
x
at every point, always. This is the reason why we need an equation in the first place, since this will provide you with the exact relationship of
y
with the values of
x
since y is a function of
x
And so, by solving for the dependent variable in an equation, meaning solving for
y
, we can relate each point
x
passes in a graph with its corresponding point with relation to the
y
But what does solving for
y
mean? In simple terms, we need to leave the variable
y
along in one side of the equality from our given equation, so that the other side of the equality contains the specific relationship (or function) telling you how
x
has to be operated on in order to obtain
y
This is something you are probably already very familiarized with. And so, we will only enter into one simple example in order to review it.
If you have difficulties at any time do not worry, the exercices below will contain other examples on graphing equations, and so, you can practice many more times this throughout out lesson.
Having the equation:
\large -x + 2y = 6
Equation 1: Linear equation to graph
y
x
to both sides and then dividing by 2 to obtain:
\large y = \frac{x}{2} + 3
Equation 2: Solving for
y
Now, setting values to the independent variable (from negative to positive), we can compute the values of y related to them and produce a table containing the ordered pairs ready to be graphed as shown below:
Table 1 and Graph 1: Graphing a single linear equation
Since this lesson is focused on solving systems of linear equations, the process of graphing linear equations as done in this section is repeated as many times as needed (once per each equation) on the system. Before continuing into solving system of equations in the next section make sure you visit the lesson on graphing linear functions using various forms to look into different techniques you can follow while making a graph for an equation. You can always select to work on problems with a different methodology than the one described here, we have selected to produce graphs this way since is the clearest and most specific method of organizing the data, but feel free to pick the way you feel most comfortable with to work, you can come back here and check your answers any time.
Now that you know how to graph equations, is time to look into solving systems by graphing. For this lesson, we will work with linear systems of two equations and two unknowns, in other words, there will be two variables (an independent =
x
, and the dependent =
y
) in each system of two equations involving both of them.
As mentioned before, this will mean that you will have to graph two equations (and thus we will have a double table for each system). So let us explain how the solutions will come up and what type of solutions can we obtain from a system like the ones we will be working with:
First, it is important to know that when solving systems by graphing, the solution is found when an ordered pair is shared between the two equations. A shared ordered pair means that when both equations are graphed, they will intersect each other at a point and this point will be the shared ordered pair, thus this ordered pair will provide the solutions for x and y that correspond to such system. Take a look at the next example for a clear picture of this:
Solve the system of equations shown below through the graphing method:
\large 2x - y = -9 \;
\large \; x + 3y = 6
Equation 3: System of linear equations
y
in each of the equations from the system:
\large 2x - y = -9 \;
\large \; x + 3y = 6
\large y = 2x + 9 \;
\large \; y = 2 - \frac{x}{3}
y
on each equation
Now let us make a table for the values of
x
y
for each of the equations:
Table 2: Values of y for each equation in the system
Graphing the system to observe their behavior:
Graph 2: Graphing each of the equations of the system
Notice from table two that both equations in the system share the ordered pair (-3, 3), if you go to the graph of both equations you can notice that the point at which intersect is this ordered pair coordinate of
x = -3
y = 3
, therefore this is the solution to our system of equations!
The rationale behind this technique is that when you graph and equation, each point in the graph represents a possible solution of this particular equation, then when having a system with more than one equation (in this case two), although each of the equations can have multiple solutions, the solution corresponding with the consistent system is that which agrees with both functions or in other words, a common point both equations share on the graph.
This takes us to realize there are two types of linear equations systems: consistent systems and inconsistent systems. Consistent systems will have at least one solution, thus, their lines on the graph will intersect at least once, while inconsistent systems of equations will not have a solution at all (the lines of the equations will never touch).
For that matter, when having a linear system consisting of two equations and two unknowns such as the ones we are working with there are three possible outcomes as you try to find their solutions:
You may find the have one solution, they may have an infinite amount of solutions or they might not have a solution at all. Therefore, the two first outcomes would correspond to a system which is consistent while the third case corresponds to an inconsistent system. And so, we conclude the linear equation system from example 1 is a consistent system since it has a solution.
When graphing systems of equations in the next exercises make sure you identify the type of system you have in each case, and the amount of solutions you have encountered.
Solve system of equations shown below by graphing:
\large 2x + 2y = 8 \;
\large \; 3x - y = 0
y
\large 2x + 2y = 8 \;
\large \; 3x - y = 0
\large y = 4 - x \;
\large \; y=3x
Equation 6: Solving for y on each equation
Now we build the table in order to find all of the ordered pairs for the system of equations. As you can see, the second expression for y is very simple to obtain making our calculations faster. After we have found their corresponding values and wrote them down in the table, we graph this equation and its companion in graph 3 below:
Table 3: Values of
y
for each equation in the system
Plotting the ordered pairs in the graph:
Thus, the system of equations found in example 2 is a consistent system with one possible solution for its
x
y
Find the solution to the system of equations shown below through the graphing method:
\large 4x + 6y = 12 \;
\large \; 2x + 3y = 9
y
\large 4x + 6y = 12 \;
\large \; 2x + 3y = 9
\large y = 2 - \frac{2}{3}x \;
\large \; y = 3 - \frac{2}{3}x
Constructing the table to obtain the ordered pairs:
y
As you can see there are no shared ordered pairs between the equations, the resultant plot is:
This means there are no solutions to this linear system of equations, and therefore, the system is inconsistent.
\large y - x = 2 \;
\large \; 4y - 8 = 4x
y
\large y - x = 2 \;
\large \; 4y - 8 = 4x
\large y = x + 2 \;
\large \; y = x + 2
Equation 10: Solving for y on each equation
Is exactly the same expression for y in both equations! Thus we dont need a double table this time since we know we will obtain the exact same results for both of them!
We will keep the double table format just for illustrative purposes so we can mark all of the ordered pairs as shared:
How to graph a linear equation when its companion in the system is passing through the exact same points? Easy! You just have to plot one, since they will overlap, thus, graph the equation of the system (pick whichever of the two) to produce:
Therefore is obvious that the linear system of equations in example 4 is consistent and by having infinite solutions.
As you can see solving systems of equations by graphing happens to be a very simple way to understand algebra from the graphic point of view. By learning how to graph an equation from a linear system and directly observe what the unique answers of the system do, we obtain a meaningful basis for algebraic expressions and how they will be useful throughout diverse and more complicated areas of mathematics.
In order to determine the number of solutions to a system of linear equations, other than by finding the slope of the lines, we can also graph the equations out and look for the intersection of the lines.
Basic Concepts: Slope equation:
m = \frac{y_2-y_1}{x_2- x_1}
, Slope intercept form: y = mx + b, Graphing linear functions using various forms, Determining number of solutions to linear equations
Related Concepts: System of linear equations, System of linear-quadratic equations, Graphing systems of linear inequalities, Graphing systems of quadratic inequalities
One Solution: Two linear equations intersect at one point. It is a consistent system of independent equation.
No Solution:Two linear equations are parallel to each other. It is an inconsistent system.
Infinite Solution:
How to find the number of solutions to systems of linear equations?
How to solve systems of linear equations by graphing?
Graph the following linear equations and find the point of intersection. State the consistency.
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Erratum: “Film Cooling Extraction Effects on the Aero-Thermal Characteristics of Rib Roughened Cooling Channel Flow” [ASME J. Turbomach., 135(2), p. 021016; DOI: 10.1115/1.4007501] | J. Turbomach. | ASME Digital Collection
This is a companion to: Film Cooling Extraction Effects on the Aero-Thermal Characteristics of Rib Roughened Cooling Channel Flow
Cukurel, B. (July 26, 2018). "Erratum: “Film Cooling Extraction Effects on the Aero-Thermal Characteristics of Rib Roughened Cooling Channel Flow” [ASME J. Turbomach., 135(2), p. 021016; DOI: 10.1115/1.4007501]." ASME. J. Turbomach. August 2018; 140(8): 087001. https://doi.org/10.1115/1.4040565
Channel flow, Cooling, Film cooling, Heat transfer, Compressors, Fluid dynamics, Gas turbines, Turbine components
There is a mistake in Eq. (5) which could misguide the reader; the correct formulation is
(Pi+Rei2/2)−(Rei+1/Rei)(Pi+1+Rei+12/2)−(SRiAR)(Ph+SRi2Rei2/2)=f¯L/DhRei2/2
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October 2021 Bridging convex and nonconvex optimization in robust PCA: Noise, outliers and missing data
Yuxin Chen,1 Jianqing Fan,2 Cong Ma,3 Yuling Yan2
1Department of Electrical and Computer Engineering, Princeton University
This paper delivers improved theoretical guarantees for the convex programming approach in low-rank matrix estimation, in the presence of (1) random noise, (2) gross sparse outliers and (3) missing data. This problem, often dubbed as robust principal component analysis (robust PCA), finds applications in various domains. Despite the wide applicability of convex relaxation, the available statistical support (particularly the stability analysis in the presence of random noise) remains highly suboptimal, which we strengthen in this paper. When the unknown matrix is well conditioned, incoherent and of constant rank, we demonstrate that a principled convex program achieves near-optimal statistical accuracy, in terms of both the Euclidean loss and the
{\ell }_{\infty }
loss. All of this happens even when nearly a constant fraction of observations are corrupted by outliers with arbitrary magnitudes. The key analysis idea lies in bridging the convex program in use and an auxiliary nonconvex optimization algorithm, and hence the title of this paper.
Y. Chen is supported in part by the AFOSR YIP award FA9550-19-1-0030, by the ONR Grant N00014-19-1-2120, by the ARO Grants W911NF-20-1-0097 and W911NF-18-1-0303, by NSF Grants CCF-1907661, IIS-1900140, IIS-2100158, and DMS-2014279 and by the Princeton SEAS innovation award. J. Fan is supported in part by the NSF Grants DMS-1662139 and DMS-1712591, the ONR Grant N00014-19-1-2120 and the NIH Grant 2R01-GM072611-14.
Author names are sorted alphabetically. Y. Chen is the corresponding author.
Yuxin Chen. Jianqing Fan. Cong Ma. Yuling Yan. "Bridging convex and nonconvex optimization in robust PCA: Noise, outliers and missing data." Ann. Statist. 49 (5) 2948 - 2971, October 2021. https://doi.org/10.1214/21-AOS2066
Received: 1 January 2020; Revised: 1 September 2020; Published: October 2021
Keywords: convex relaxation , ℓ∞ guarantees , leave-one-out analysis , Robust principal component analysis
Yuxin Chen, Jianqing Fan, Cong Ma, Yuling Yan "Bridging convex and nonconvex optimization in robust PCA: Noise, outliers and missing data," The Annals of Statistics, Ann. Statist. 49(5), 2948-2971, (October 2021)
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Documentation Talk:Reference Section 4 - POV-Wiki
Documentation Talk:Reference Section 4
Blob Formula
{\displaystyle {density}={strength}\cdot \left(1-\left({\frac {\min(distance,radius)}{radius}}\right)^{2}\right)^{2}}
Better Blob Formula
{\displaystyle {\textit {density}}={\textit {strength}}\cdot \left(1-\left({\frac {\min({\textit {distance}},{\textit {radius}})}{\textit {radius}}}\right)^{2}\right)^{2}}
inline pictures rendered from the portfolio
I know it might seems childish or simple, but even if the graphics illustrations in 2D are nice to explain the measurement (so, keep them), it might be more than a good move to also provide for each shape a sample of render (and we already have them, in the portfolio of povray, section allobjects).
It is pretty obvious what a sphere is, but even the torus had to rely on the donut naming to give everyone a clue. Text is good, but a really rendered shape for each entry (and whose source could be found in the distribution) might really enhance that section.
Actual changes to the pages would be: upload the rendered images from portfolio, add the pictures in the pages.
Some better images could be from the window's povray menu/Insert
The issue of new images is something that's been raised before (other sections) ... it's on my list but low priority at the moment --jholsenback 17:30, 24 October 2010 (UTC)
Update the syntax for ovus (at the beginning of the page, in the object chapter)
--Le Forgeron 08:36, 3 December 2010 (UTC)
done: --jholsenback 13:40, 3 December 2010 (UTC)
FINITE_SOLID_OBJECT | FINITE_PATCH_OBJECT |
INFINITE_SOLID_OBJECT | ISOSURFACE_OBJECT | PARAMETRIC_OBJECT |
CSG_OBJECT | LIGHT_SOURCE |
object { OBJECT_IDENTIFIER [OBJECT_MODIFIERS...] }
BLOB | BOX | CONE | CYLINDER | HEIGHT_FIELD | JULIA_FRACTAL |
LATHE | OVUS | PRISM | SPHERE | SPHERESWEEP | SUPERELLIPSOID |
SOR | TEXT | TORUS
FINITE_PATCH_OBJECT:
BICUBIC_PATCH | DISC | MESH | MESH2 | POLYGON | TRIANGLE |
INFINITE_SOLID_OBJECT:
PLANE | POLY | CUBIC | QUARTIC | QUADRIC
ISOSURFACE_OBJECT:
PARAMETRIC_OBJECT:
Retrieved from "https://wiki.povray.org/content?title=Documentation_Talk:Reference_Section_4&oldid=3309"
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Lori has written the conjectures below. For each one, decide if it is true or not. If you believe it is not true, find a counterexample (an example that proves that the statement is false).
If a shape has four equal sides, it cannot be a parallelogram.
False. A rhombus and a square are counterexamples because both have
4
equal sides and opposite sides are parallel like in all parallelograms.
θ
1
θ
45^\circ
Look at the Math Notes box for Lesson 4.1.4 for extra help.
\large\frac{\text{opp}}{\text{adj}}
1
What if the lines are not parallel? Draw a diagram.
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Acyclic covers | EMS Press
Acyclic covers
The following question, which is directly related to the Whitehead problem of subcomplexes of acyclic 2-complexes, is studied: If
\frak P
is a class of groups, X is a 2-dimensional CW-complex and X' is an acyclic, infinite cyclic cover of X with
\pi_1(X')
\frak P
, must X' be contractible? A positive answer is given if X is finite and
\frak P
is the class of amenable groups.
John J. Millson, Acyclic covers. Comment. Math. Helv. 74 (1999), no. 2, pp. 173–178
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When parallel lines are cut by transversals | StudyPug
What are parallel lines? They are lines that are equidistant (the same distance apart) from each other but never meet. They also point in the same direction.
When a set of parallel lines gets cut by another line, that other line is called a transversal. Having this transversal line paves the way for a variety of helpful information that we can derive from the parallel lines' angles.
When we've got a line with a transversal through it, we can identify corresponding angles from the two parallel lines that are the same. Therefore, corresponding angles are equal. This is the corresponding angles definition. We'll have a chart below to summarize this and the following concept at the end with visual depictions of what these terms mean.
Alternate interior angles are found on the opposite sides of the transversal but inside the two lines. These angles are on the interior of the lines that cross, but on alternate sides of the transversal. This is the alternate interior angles definition.
Here's a summary chart of the above concepts, plus a few more pairs of angles that come about due to parallel lines and a transversal. Unlike regular pairs of lines and angles, these ones have special rules to their angles based off the parallel lines. They'll be very useful in helping you solve questions related to pairs of angles.
Find the missing values and then explain your reasoning.
a) angle 6
\cong
b) angle 2
\cong
c) if m angle 8 = 60°, m angle 1 = ?
d) angle 3 and angle ? are supplementary
\cong
Angle 6 is congruent to angle 2 due to the corresponding angle rule.
Angle 6 is also congruent to angle 7 due to vertical angle congruence.
Lastly, angle 6 is congruent to angle 3 because of the corresponding angle rule between angle 3 and angle 7.
\cong
Angle 2 is equalled to angle 6 since they're corresponding angles.
Lastly, angle 2 is congruent to angle 7 based on the corresponding angle rule or alternate exterior angle theorem.
We get that m angle 1 = 60°. This is based on the alternate exterior angle theorem.
Supplementary means that the sum the two angles = 180°
So angle 3 + angle 5 is supplementary due to the consecutive interior angle theorem.
In addition, angle 8, angle 1, angle 4 are also supplementary to angle 3.
Angle 130° = 2x°
Given: AB||CD
Prove: angle 1
\cong
angle 2
\cong
Vertical angle congruence theorem
\cong
Corresponding angle postulate
\cong
Take a look at this interactive online diagram to help you see the angle relationships that exist when there is a set of parallel lines and a transversal.
To move further into this concept, learn more about parallel line equation and how to deal with a combination of parallel and perpendicular line equations.
Basic Concepts: Pairs of lines and angles
Related Concepts: Parallel line equation, Combination of both parallel and perpendicular line equations
\angle 6
\cong
\angle 2
\cong
\angle 8
= 60°, m
\angle 1
\angle 3
\angle ?
Are supplementary
\overline {AB}
\parallel
\overline {CD}
\angle 1
\cong
\angle 2
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How to apply quadratic functions | StudyPug
y = ax^2 + bx+c
a(x-p)^2 + q
y = ax^2 + bx+c
a(x-p)^2 + q
y = ax^2 + bx+c
a(x-p)^2 + q
Related Concepts: Applications of quadratic equations, Factor theorem
Different Ways to Solve Quadratic Functions
A farmer wants to build a rectangular pig farm beside a river. He has 200 meters of fencing material, and there is no need for fencing on the side along the river. What are the dimensions of the largest pig farm this farmer can build?
The sum of two integers is 10, and the product is a maximum. Find the integers.
The sum of two integers is 10, and the sum of their squares is a minimum. Find the integers.
John stands on the roof of a building and throws a ball upwards. The ball's height above the ground is given by the formula:
h = - 3{t^2} + 12t + 15
, where h is the height in meters at t seconds after the ball is thrown.
How high is John above the ground when he throws the ball?
Find the height above the ground of the ball 1 second after the ball is thrown.
How long does it take for the ball to reach the maximum height?
Find the maximum height above the ground reached by the ball.
A school Halloween dance charges $5 for admission, and 200 students are willing to attend the dance. For every 25 cents increase in price, attendance drops by 4 students.
What price should the school charge to maximize the revenue?
How many students would need to attend the dance in order to generate the maximum revenue?
What is the maximum revenue?
y = ax^2 + bx+c
a(x-p)^2 + q
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For each question below, describe a realistic method to collect information to answer the question. Be sure to indicate whether you would use a sample or census and whether your results would be parameters or statistics.
What percentage of American League baseball players had a batting average above
.300
this season?
This information could be found on the web for all American League players. It would be a census, and the answer would be a parameter.
How much pressure can be exerted on a chicken egg before it breaks?
An experiment would need to be conducted on a sample of eggs. The findings would be a statistic.
How many hours of television does a high school student watch per day?
Random high school students could be surveyed, possibly from different high schools in different parts of the country. Surveying every high school student would be almost impossible, so this survey would be a sample, and the answer would be a statistic.
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Commercially available concentrated Hcl contains 38% Hcl by mass (1) what is the molarity of the solution ( density - Chemistry - Some Basic Concepts of Chemistry - 7801845 | Meritnation.com
Commercially available concentrated Hcl contains 38% Hcl by mass. (1) what is the molarity of the solution ( density of solution =1.19 g mL -1)
(2) What volume of concentrated Hcl is required is required to make 1.0 L of an 0.10 M Hcl.
A 38% by mass solution of HCl means that 38 g of HCl is present in 62g of water, thereby making the total mass of solution equal to 100g. The molar mass of HCl is 36.5g. Therefore, the number of moles of HCl that are present in 38g of HCl are
= mass of HCl / molar mass of HCl
We known, Molarity = Number of moles/ Volume of solution (in L)
As Volume of solutions is not known therefore, we will take help of density
Volume = 100/ 1.1
On substituting the above values in the formula of molarity, we will get
×
1000/ 90.91
The molarity of 38 % solution is 11.44 M
As we are supposed to make it 0.10 M in 1 L, therefore we can use the formula
M1V1 = M2V2 ...(ii)
Where 1 stands for commercially available HCl and 2 stands for final solution.
on substituing the above value in equation (ii), we will get
V1 = 0.10
×
1 / 11.44
or 8.74 mL of HCl would be required.
1) Concentrated HCl 38% by mass i.e 100gm of HCL soclution contains 38gm of HCl moles of HCl = given mass/molar mass = 38/36.5 Now density of solution = 1.19 g mL-1Since solution is 100gm. Density = Mass/volume = Volume= mass/density = 100/1.19 ml Molarity = Moles of Solute/volume of Solution in L M = (38/36.5)*(1000*1.19/100) = 12.38 M
2) M = Given Mass/Molecular mass* volume of solution in L 0.10M = x/36.5*1 = 0.10M * 36.5 = x = x = 3.65 gm Moles of HCl = 3.65/36.5 = 0.1 mole at STP 1mole = 22.4 L, so 0.1 mole = 2.24 L
Ans1 = 12.38M
Ans2 = 2.24 L
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Basic Concepts: Probability of independent events, Comparing experimental and theoretical probability, Simplifying rational expressions and restrictions, Solving rational equations
Related Concepts: Radius and interval of convergence with power series, Taylor and maclaurin series
\frac{{7!}}{{5!}}
\frac{{\left( {n + 3} \right)!}}{{n!}}
\frac{{\left( {n - 1} \right)!}}{{\left( {n + 2} \right)!}}
\frac{{\left( {n + 1} \right)!\;\;\left( {n - 3} \right)!}}{{{{(n!)}^2}}}
\frac{{n!}}{{\left( {n - 2} \right)!\;\;3!}} = 7
\frac{{n!}}{{\left( {{1^{st}}\;repetition} \right)!\;\;\;\left( {{2^{nd}}\;repetition} \right)!\;\;\;\left( {{3^{rd}}\;repetition} \right)!\; \ldots ..}}
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The double dispersion operator in backscattering: Hölder estimates and optimal Sobolev estimates for radial potentials | EMS Press
We study the problem of recovering the singularities of a potential
q
from backscattering data. In particular, we prove two new different estimates for the double dispersion operator
Q_2
of backscattering, the first nonlinear term in the Born series. In the first, by measuring the regularity in the Hölder scale, we show that there is a one derivative gain in the integrablity sense for suitably decaying potentials
q\in W^{\beta,2}(\mathbb{R}^n)
\beta \ge (n-2)/2
n \ge 3
. In the second, we give optimal estimates in the Sobolev scale for
Q_2(q)
n\ge 2
q
is radial. In dimensions 2 and 3 this result implies an optimal result of recovery of singularities from the Born approximation.
Cristóbal J. Meroño, The double dispersion operator in backscattering: Hölder estimates and optimal Sobolev estimates for radial potentials. Rev. Mat. Iberoam. 37 (2021), no. 3, pp. 1175–1205
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Guppy Multiple Moving Average (GMMA) Definition
What Does the GMMA Tell You?
GMMA vs. EMA
The Guppy Multiple Moving Average (GMMA) is a technical indicator that aims to anticipate a potential breakout in the price of an asset. The term gets its name from Daryl Guppy, an Australian financial columnist and book author who developed the concept in his book, "Trading Tactics."
The GMMA uses the exponential moving average (EMA) to capture the difference between price and value in a stock. A convergence in these factors is associated with a significant trend change. Guppy maintains that the GMMA is not a lagging indicator but a prior warning of a developing change in price and value.
Guppy Multiple Moving Average (GMMA) Formula and Calculation
The formula for the Guppy indicator uses exponential moving averages (EMA). There is a short-term group of MAs and a long-term group of MAs, both containing six MAs, for a total of 12. However, one can insert their preferred number of periods, N, into the calculation to find each of the MA values.
\begin{aligned} &EMA = \left[\text{Close price} - EMA_{previous}\right]*M+EMA_{previous}\\ &\textbf{or:}\\ &SMA = \frac{\text{Sum of } N \text{ closing prices}}{N}\\ \\ \\ &\textbf{where:}\\ &EMA = \text{exponential moving average}\\ &EMA_{previous} = \text{the exponential moving average from the previous period}\\ &\text{(The } SMA \text{ can substitute for the } EMA_{previous} \text{ for the first calculation)}\\ &\text{Multiplier } M = \frac{2}{N+1}\\ &SMA = \text{simple moving average}\\ &N = \text{number of periods}\\ \end{aligned}
EMA=[Close price−EMAprevious]∗M+EMApreviousor:SMA=NSum of N closing priceswhere:EMA=exponential moving averageEMAprevious=the exponential moving average from the previous period(The SMA can substitute for the EMAprevious for the first calculation)Multiplier M=N+12SMA=simple moving averageN=number of periods
Repeat the steps below for each of the required MAs. Alter the N value to calculate the EMA you want. For example, use three to calculate the three-period average, and use 60 to calculate the 60-period EMA.
Calculate the SMA for N.
Calculate the multiplier using the same N value.
Use the most recent closing price, the multiplier, and SMA to calculate the EMA. The SMA is placed in the EMA previous day spot in the calculation. Once the EMA has been calculated, the SMA is no longer needed since the EMA calculation can be used in the EMA previous day spot for the next calculation.
Repeat the process for the next N value, until you have the EMA reading for all 12 MAs.
The degree of separation between the short- and long-term MAs can be used as an indicator of trend strength. If there's a wide separation, then the prevailing trend is strong. Narrow separation, or lines that are crisscrossings, on the other hand, indicates a weakening trend or a period of consolidation.
The crossover of the short- and long-term MAs represent trend reversals. If the short-term crosses above the long-term MAs, then a bullish reversal has occurred. Conversely, if the short-term MAs cross below the longer-term ones, then a bearish reversal is occurring.
Meanwhile, when both groups of MAs are moving horizontally, or mostly moving sideways and heavily intertwined, it means the asset lacks a price trend, and therefore may not be a good candidate for trend trades. These periods may be good for range trading, though.
The GMMA can be employed to identify changes in trends or gauge the strength of the current trend and are best used in conjunction with other technical indicators.
The indicator can also be used for trade signals. When the short-term group passes above the long-term group of MAs, buy. When the short-term group passes below the longer-term group, sell. These signals should be avoided when the price and the MAs are moving sideways. Following a consolidation period, watch for a crossover and separation. When the lines start to separate this often means a breakout from the consolidation has occurred and a new trend could be underway.
During a strong uptrend, when the short-term MAs move back toward the longer-term MAs (but don't cross) and then start to move back to the upside, this is another opportunity to enter into long trades in the trending direction. The same concept applies to downtrends for entering short trades.
The Guppy Multiple Moving Average (GMMA) vs. an Exponential Moving Average (EMA)
The GMMA is composed of 12 EMAs, so it is essentially the same thing as an EMA. The Guppy is a collection of EMAs that the creator believed helped isolate trades, spot opportunities, and warn about price reversals.
The multiple lines of the Guppy help some traders see the strength or weakness in a trend better than if only using one or two EMAs.
Limitations of the GMMA
The main limitation of the Guppy, and the EMAs it is composed of, is that it is a lagging indicator. Each EMA represents the average price from the past. It does not predict the future.
Waiting for the averages to crossover can at times mean an entry or exit that is far too late, as the price has already moved aggressively. All MAs are also prone to whipsaws. This is when there is a crossover, potentially resulting in a trade, but the price doesn't move as expected and then the averages cross again resulting in a loss.
Traders should use the GMMA in conjunction with other technical indicators to maximize their odds of success. For example, traders might look at the relative strength index (RSI) to confirm whether a trend is getting top-heavy and poised for a reversal, or look at various chart patterns to determine other entry or exit points after a GMMA crossover.
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interval - Maple Help
Home : Support : Online Help : Programming : Logic : Boolean : verify : interval
verify that the first argument is a point in the interval
verify(expr1, expr2, interval(character))
anything, assumed to be of type range(algebraic)
(optional) description of the character of the interval
An interval is either an open, closed, or semi-open contiguous subset of the real line.
The verify(expr1, expr2, interval(character)) calling sequence returns true if it can determine that the point expr1 is located in the interval defined by the second argument.
The options available for describing the character of the interval are: open..open, closed..closed, open..closed, and closed..open. The abbreviations open and closed are equivalent to open..open and closed..closed, respectively.
This verification is a short form for calling the appropriate forms of the verifications greater_than, greater_equal, less_than, and less_equal.
If expr1 is not of type algebraic or expr2 is not of type range(algebraic), then false is returned.
\mathrm{verify}\left(\mathrm{\pi },3..4,'\mathrm{interval}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{verify}\left(\mathrm{\pi },\mathrm{\pi }..4,'\mathrm{interval}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{verify}\left(\mathrm{\pi },\mathrm{\pi }..4,'\mathrm{interval}'\left('\mathrm{closed}'\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{verify}\left([4,4],[3..5,4..5],'[\mathrm{interval},\mathrm{interval}]'\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{verify}\left([a,b],[3..5,4..5],'[\mathrm{interval},\mathrm{interval}]'\right)
\textcolor[rgb]{0,0,1}{\mathrm{FAIL}}
\mathrm{verify}\left([a,b],[3..5,4..5],'\mathrm{truefalse}\left([\mathrm{interval},\mathrm{interval}]\right)'\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{verify}\left(\mathrm{Vector}\left([\mathrm{exp}\left(1\right),\mathrm{exp}\left(2\right)]\right),\mathrm{Vector}\left([2..5,4..12]\right),'\mathrm{truefalse}\left(\mathrm{Vector}\left(\mathrm{interval}\right)\right)'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
Given a list of coordinates, the following selects all those that fall in the interval
0.05..0.15
\mathrm{select}\left('\mathrm{verify}',[[0.,0.],[0.065391,0.065345],[0.12229,0.12198],[0.18627,0.18520]],[0.05..0.15,'\mathrm{dummy}'],'[\mathrm{interval},\mathrm{true}]'\right)
[[\textcolor[rgb]{0,0,1}{0.065391}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.065345}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{0.12229}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.12198}]]
verify/less_than
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Outermorphism - Wikipedia
Unital algebra homomorphism of exterior algebras
This article is about outermorphisms of geometric algebra. For the concept in group theory, see outer automorphism group.
In geometric algebra, the outermorphism of a linear function between vector spaces is a natural extension of the map to arbitrary multivectors.[1] It is the unique unital algebra homomorphism of exterior algebras whose restriction to the vector spaces is the original function.[a]
{\displaystyle f}
{\displaystyle \mathbb {R} }
-linear map from
{\displaystyle V}
{\displaystyle W}
. The extension o{\displaystyle f}
to an outermorphism is the unique map
{\displaystyle \textstyle {\underline {\mathsf {f}}}:\bigwedge (V)\to \bigwedge (W)}
{\displaystyle {\underline {\mathsf {f}}}(1)=1}
{\displaystyle {\underline {\mathsf {f}}}(x)=f(x)}
{\displaystyle {\underline {\mathsf {f}}}(A\wedge B)={\underline {\mathsf {f}}}(A)\wedge {\underline {\mathsf {f}}}(B)}
{\displaystyle {\underline {\mathsf {f}}}(A+B)={\underline {\mathsf {f}}}(A)+{\underline {\mathsf {f}}}(B)}
{\displaystyle x}
and all multivectors
{\displaystyle A}
{\displaystyle B}
{\displaystyle \textstyle \bigwedge (V)}
denotes the exterior algebra over
{\displaystyle V}
. That is, an outermorphism is a unital algebra homomorphism between exterior algebras.
The outermorphism inherits linearity properties of the original linear map. For example, we see that for scalars
{\displaystyle \alpha }
{\displaystyle \beta }
{\displaystyle x}
{\displaystyle y}
{\displaystyle z}
, the outermorphism is linear over bivectors:
{\displaystyle {\begin{aligned}{\underline {\mathsf {f}}}(\alpha x\wedge z+\beta y\wedge z)&={\underline {\mathsf {f}}}((\alpha x+\beta y)\wedge z)\\[6pt]&=f(\alpha x+\beta y)\wedge f(z)\\[6pt]&=(\alpha f(x)+\beta f(y))\wedge f(z)\\[6pt]&=\alpha (f(x)\wedge f(z))+\beta (f(y)\wedge f(z))\\[6pt]&=\alpha \,{\underline {\mathsf {f}}}(x\wedge z)+\beta \,{\underline {\mathsf {f}}}(y\wedge z),\end{aligned}}}
which extends through the axiom of distributivity over addition above to linearity over all multivectors.
Adjoint[edit]
{\displaystyle {\underline {\mathsf {f}}}}
be an outermorphism. We define the adjoint of
{\displaystyle {\overline {\mathsf {f}}}}
to be the outermorphism that satisfies the property
{\displaystyle {\overline {\mathsf {f}}}(a)\cdot b=a\cdot {\underline {\mathsf {f}}}(b)}
{\displaystyle a}nd
{\displaystyle b}
{\displaystyle \cdot }
is the nondegenerate symmetric bilinear form (scalar product of vectors).
This results in the property that
{\displaystyle {\overline {\mathsf {f}}}(A)*B=A*{\underline {\mathsf {f}}}(B)}
for all multivectors
{\displaystyle A}
{\displaystyle B}
{\displaystyle *}
is the scalar product of multivectors.
If geometric calculus is available, then the adjoint may be extracted more directly:
{\displaystyle {\overline {\mathsf {f}}}(a)=\nabla _{b}\left\langle a{\underline {\mathsf {f}}}(b)\right\rangle .}
The above definition of adjoint is like the definition of the transpose in matrix theory. When the context is clear, the underline below the function is often omitted.
It follows from the definition at the beginning that the outermorphism of a multivector
{\displaystyle A}
is grade-preserving:[2]
{\displaystyle {\underline {\mathsf {f}}}(\left\langle A\right\rangle _{r})=\left\langle {\underline {\mathsf {f}}}(A)\right\rangle _{r}}
{\displaystyle \langle ~\rangle _{r}}
indicates the
{\displaystyle r}
-vector part of
{\displaystyle A}
Since any vector
{\displaystyle x}
{\displaystyle x=1\wedge x}
, it follows that scalars are unaffected with
{\displaystyle {\underline {\mathsf {f}}}(1)=1}
.[b] Similarly, since there is only one pseudoscalar up to a scalar multiplier, we must have
{\displaystyle {\underline {\mathsf {f}}}(I)\propto I}
. The determinant is defined to be the proportionality factor:[3]
{\displaystyle \det {\mathsf {f}}={\underline {\mathsf {f}}}(I)I^{-1}}
The underline is not necessary in this context because the determinant of a function is the same as the determinant of its adjoint. The determinant of the composition of functions is the product of the determinants:
{\displaystyle \det({\mathsf {f}}\circ {\mathsf {g}})=\det {\mathsf {f}}\det {\mathsf {g}}}
If the determinant of a function is nonzero, then the function has an inverse given by
{\displaystyle {\underline {\mathsf {f}}}^{-1}(X)={\frac {{\overline {\mathsf {f}}}(XI)I^{-1}}{\det {\mathsf {f}}}}={\overline {\mathsf {f}}}(XI)[{\overline {\mathsf {f}}}(I)]^{-1},}
and so does its adjoint, with
{\displaystyle {\overline {\mathsf {f}}}^{-1}(X)={\frac {I^{-1}{\underline {\mathsf {f}}}(IX)}{\det {\mathsf {f}}}}=[{\underline {\mathsf {f}}}(I)]^{-1}{\underline {\mathsf {f}}}(IX).}
The concepts of eigenvalues and eigenvectors may be generalized to outermorphisms. Let
{\displaystyle \lambda }
{\displaystyle B}
be a (nonzero) blade of grade
{\displaystyle r}
. We say that a
{\displaystyle B}
is an eigenblade of the function with eigenvalue
{\displaystyle \lambda }
if[4]
{\displaystyle {\underline {\mathsf {f}}}(B)=\lambda B.}
It may seem strange to consider only real eigenvalues, since in linear algebra the eigenvalues of a matrix with all real entries can have complex eigenvalues. In geometric algebra, however, the blades of different grades can exhibit a complex structure. Since both vectors and pseudovectors can act as eigenblades, they may each have a set of eigenvalues matching the degrees of freedom of the complex eigenvalues that would be found in ordinary linear algebra.
The identity map and the scalar projection operator are outermorphisms.
A rotation of a vector by a rotor
{\displaystyle R}
{\displaystyle f(x)=RxR^{-1}}
with outermorphism
{\displaystyle {\underline {\mathsf {f}}}(X)=RXR^{-1}.}
We check that this is the correct form of the outermorphism. Since rotations are built from the geometric product, which has the distributive property, they must be linear. To see that rotations are also outermorphisms, we recall that rotations preserve angles between vectors:[5]
{\displaystyle x\cdot y=(RxR^{-1})\cdot (RyR^{-1})}
Next, we try inputting a higher grade element and check that it is consistent with the original rotation for vectors:
{\displaystyle {\begin{aligned}{\underline {\mathsf {f}}}(x\wedge y)&=R(x\wedge y)R^{-1}\\&=R(xy-x\cdot y)R^{-1}\\&=RxyR^{-1}-R(x\cdot y)R^{-1}\\&=RxR^{-1}RyR^{-1}-x\cdot y\\&=(RxR^{-1})\wedge (RyR^{-1})+(RxR^{-1})\cdot (RyR^{-1})-x\cdot y\\&=(RxR^{-1})\wedge (RyR^{-1})+x\cdot y-x\cdot y\\&=f(x)\wedge f(y)\end{aligned}}}
The orthogonal projection operator
{\displaystyle {\mathcal {P}}_{B}}
onto a blade
{\displaystyle B}
is an outermorphism:
{\displaystyle {\mathcal {P}}_{B}(x\wedge y)={\mathcal {P}}_{B}(x)\wedge {\mathcal {P}}_{B}(y).}
Nonexample – orthogonal rejection operator
In contrast to the orthogonal projection operator, the orthogonal rejection
{\displaystyle {\mathcal {P}}_{B}^{\perp }}
by a blade
{\displaystyle B}
is linear but is not an outermorphism:
{\displaystyle {\mathcal {P}}_{B}^{\perp }(1)=1-{\mathcal {P}}_{B}(1)=0\neq 1.}
Nonexample – grade projection operator
An example of a multivector-valued function of multivectors that is linear but is not an outermorphism is grade projection where the grade is nonzero, for example projection onto grade 1:
{\displaystyle \langle x\wedge y\rangle _{1}=0}
{\displaystyle \langle x\rangle _{1}\wedge \langle y\rangle _{1}=x\wedge y}
^ See particularly Exterior algebra § Functoriality.
^ Except for the case where
{\displaystyle f}
is the zero map, when it is required by axiom.
^ Dorst, Doran & Lasenby 2001.
^ Hestenes & Sobczyk 1987, p. 68. (here at Google Books)
^ Perwass 2008.
Hestenes, D.; Sobczyk, G. (1987), Clifford Algebra to Geometric Calculus: A Unified Language for Mathematics and Physics, Fundamental Theories of Physics, vol. 5, Springer, ISBN 90-277-2561-6
Crumeyrolle, A.; Ablamowicz, R.; Lounesto, P. (1995), Clifford Algebras and Spinor Structures: A Special Volume Dedicated to the Memory of Albert Crumeyrolle (1919–1992), Mathematics and Its Applications, vol. 321, Springer, p. 105, ISBN 0-7923-3366-7
Baylis, W.E. (1996), Clifford (Geometric) Algebras: With Applications in Physics, Mathematics, and Engineering, Springer, p. 71, ISBN 0-8176-3868-7
Dorst, L.; Doran, C.J.L.; Lasenby, J. (2001), Applications of geometric algebra in computer science and engineering, Springer, p. 61, ISBN 0-8176-4267-6
D'Orangeville, C.; Anthony, A.; Lasenby, N. (2003), Geometric Algebra For Physicists, Cambridge University Press, p. 343, ISBN 0-521-48022-1
Perwass, C. (2008), Geometric Algebra with Applications in Engineering, Geometry and Computing, vol. 4, Springer, p. 23, ISBN 3-540-89067-X
Joot, P. (2014), Exploring physics with Geometric Algebra, p. 157
Retrieved from "https://en.wikipedia.org/w/index.php?title=Outermorphism&oldid=1032661612"
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The geometry of $E$-manifolds | EMS Press
E
Universitat Politècnica de Catalunya, Barcelona, Spain and Sorbonne Université, Paris, France
Motivated by the study of symplectic Lie algebroids, we focus on a type of algebroid (called an
E
-tangent bundle) which is particularly well-suited to the study of singular differential forms and their cohomology. This setting generalizes
b
-symplectic manifolds, foliated manifolds, and a wide class of Poisson manifolds. We generalize Moser’s theorem to this setting, and use it to construct symplectomorphisms between singular symplectic forms. We give applications of this machinery (including the study of Poisson cohomology), and study specific examples of a few of them in depth.
Eva Miranda, Geoffrey Scott, The geometry of
E
-manifolds. Rev. Mat. Iberoam. 37 (2021), no. 3, pp. 1207–1224
|
A\mathrm{sin}\left(x\right)
A
x
t
n
p
p
p
The trace=n option specifies that a number of previous frames of the animation be kept visible. When
n
n+1
n=5
When
is a list of integers, then the frames in those positions are the frames that remain visible. Each integer in
n=0
\mathrm{with}\left(\mathrm{plots}\right):
\mathrm{animate}\left(\mathrm{plot},[A{x}^{2},x=-4..4],A=-3..3\right)
\mathrm{animate}\left(\mathrm{plot},[A{x}^{2},x=-4..4],A=-3..3,\mathrm{trace}=5,\mathrm{frames}=50\right)
\mathrm{animate}\left(\mathrm{plot},[A{x}^{2},x=-4..4],A=-3..3,\mathrm{trace}=[30,35,40,45,50],\mathrm{frames}=50\right)
\mathrm{animate}\left(\mathrm{plot3d},[A\left({x}^{2}+{y}^{2}\right),x=-3..3,y=-3..3],A=-2..2,\mathrm{style}=\mathrm{patchcontour}\right)
\mathrm{animate}\left(\mathrm{implicitplot},[{x}^{2}+{y}^{2}={r}^{2},x=-3..3,y=-3..3],r=1..3,\mathrm{scaling}=\mathrm{constrained}\right)
\mathrm{animate}\left(\mathrm{implicitplot},[{x}^{2}+Axy-{y}^{2}=1,x=-2..2,y=-3..3],A=-2..2,\mathrm{scaling}=\mathrm{constrained}\right)
\mathrm{animate}\left(\mathrm{plot},[[\mathrm{sin}\left(t\right),\mathrm{sin}\left(t\right)\mathrm{exp}\left(-\frac{t}{5}\right)],t=0..x],x=0..6\mathrm{\pi },\mathrm{frames}=50\right)
\mathrm{animate}\left(\mathrm{plot},[[\mathrm{cos}\left(t\right),\mathrm{sin}\left(t\right),t=0..A]],A=0..2\mathrm{\pi },\mathrm{scaling}=\mathrm{constrained},\mathrm{frames}=50\right)
\mathrm{animate}\left(\mathrm{plot},[[\frac{1-{t}^{2}}{1+{t}^{2}},\frac{2t}{1+{t}^{2}},t=-10..A]],A=-10..10,\mathrm{scaling}=\mathrm{constrained},\mathrm{frames}=50,\mathrm{view}=[-1..1,-1..1]\right)
\mathrm{opts}≔\mathrm{thickness}=5,\mathrm{numpoints}=100,\mathrm{color}=\mathrm{black}:
\mathrm{animate}\left(\mathrm{spacecurve},[[\mathrm{cos}\left(t\right),\mathrm{sin}\left(t\right),\left(2+\mathrm{sin}\left(A\right)\right)t],t=0..20,\mathrm{opts}],A=0..2\mathrm{\pi }\right)
B≔\mathrm{plot3d}\left(1-{x}^{2}-{y}^{2},x=-1..1,y=-1..1,\mathrm{style}=\mathrm{patchcontour}\right):
\mathrm{opts}≔\mathrm{thickness}=5,\mathrm{color}=\mathrm{black}:
\mathrm{animate}\left(\mathrm{spacecurve},[[t,t,1-2{t}^{2}],t=-1..A,\mathrm{opts}],A=-1..1,\mathrm{frames}=11,\mathrm{background}=B\right)
\mathrm{animate}\left(\mathrm{ball},[0,\mathrm{sin}\left(t\right)],t=0..4\mathrm{\pi },\mathrm{scaling}=\mathrm{constrained},\mathrm{frames}=100\right)
\mathrm{sinewave}≔\mathrm{plot}\left(\mathrm{sin}\left(x\right),x=0..4\mathrm{\pi }\right):
\mathrm{animate}\left(\mathrm{ball},[t,\mathrm{sin}\left(t\right)],t=0..4\mathrm{\pi },\mathrm{frames}=50,\mathrm{background}=\mathrm{sinewave},\mathrm{scaling}=\mathrm{constrained}\right)
\mathrm{animate}\left(\mathrm{ball},[t,\mathrm{sin}\left(t\right)],t=0..4\mathrm{\pi },\mathrm{frames}=50,\mathrm{trace}=10,\mathrm{scaling}=\mathrm{constrained}\right)
\mathrm{animate}\left(F,[\mathrm{\theta }],\mathrm{\theta }=0..2\mathrm{\pi },\mathrm{background}=\mathrm{plot}\left([\mathrm{cos}\left(t\right)-2,\mathrm{sin}\left(t\right),t=0..2\mathrm{\pi }]\right),\mathrm{scaling}=\mathrm{constrained},\mathrm{axes}=\mathrm{none}\right)
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Learn to multiply a binomial by a binomial | StudyPug
Solving problems of multiplying binomial by binomial requires some skills. Let's learn how to do it by using the FOIL method in this lesson.
Basic Concepts: Product rule of exponents, Multiplying monomial by monomial, Multiplying monomial by binomial
Related Concepts: Solving polynomials with unknown coefficients, Solving polynomials with unknown constant terms, Factoring polynomials:
x^2 + bx + c
, Solving polynomials with the unknown "b" from
ax^2 + bx + c
{(x^3-4)(2x+3)}
{(x^2-3)(4x-7)}
|
Solve linear equations with distributive property | StudyPug
a(x + b) = c
Solving two-step linear equations using distributive property:
\;a\left( {x + b} \right) = c
ax + b = c
{x \over a} + b = c
\;a\left( {x + b} \right) = c
ax + b = c
{x \over a} + b = c
What is the definition of distributive property? Sometimes called the distributive law of multiplication and division, distributive property helps us solve equations like a(b+c). In distributive property, you'll have to take the numbers from inside the parentheses out (factor them out) by multiplying what's outside with the terms inside.
In order to tackle the distributive property, first recognize the question you're given and whether distributive property is the appropriate method to use.
Let's take a(b+c). In this case, if b+c can no longer be simplified (remember you're supposed to solve everything inside parentheses first), then you'll take "a" and multiply it with the two different terms inside the parentheses, which consists of "+b" and "+c". Take note of the signs in front of the terms. Multiplying in "a" with these two terms gives us:
If it was a(b-c), we'll be multiplying a with the terms "+b" and "-c". This will give us:
a(b-c)=ab-ac
Be mindful of the sign so that you don't end up with the wrong answer. Here's a reference to a color coded explanation of the distributive property if you need to visualize the concept.
Let's try solving equations using distributive property. We'll guide you through two ways that the following question can be solved.
Question: 11.32 = 8(0.61 + x)
11.32 = 8(0.61+x)
Divide both side by 8, you get 1.415 = 0.61 +x
Minus both side by 0.61, 0.805 = x
In this solution, we first try to move all the non-x terms over to one side. The first step of dividing both sides by 8 helps us move the 8 from the right side to the left side. This allows us to get rid of the parentheses too as the terms no longer have to be multiplied out (there's nothing to multiply them with). Then, in order to move 0.61 to the left side, we'll minus it from both sides. In the end, we'll get that x ultimately equals 0.805.
Open up the right side, you get 11.32 = 4.88+8x
Minus both side by 4.88, you get 6.44 = 8x
Divide both side by 8, you get 0.805 = x
In solution 2, we're truly using the distributive property. We're going to distribute 8 into the terms in the parentheses, which are "+0.61" and "+x". This gives us 4.88 + 8x on the right side of the equation. Then we can move over 4.88 to the left side with the rest of the non-x terms by minusing it from both sides. Lastly, divide both sides by 8 to move the 8 over to the left side and you'll find that x = 0.805. Both methods will bring you to the same correct answer.
Try out more distributive property examples on StudyPug that can help you firmly grasp the concept before your next test. If you want more questions, feel free to look online for a distributive property worksheet and apply the same method we've shown you to solve the problems.
ax + b = c
{x \over a} + b = c
a(x + b) = c
a(x + b) = c
When you see equations in the form of a(x+b) =c, you can remove the bracket and rewrite the equations into ax+ab =c using the distributive property. In this lesson, we will make use of this property to help us solve linear equations.
Basic Concepts: Solving two-step linear equations using distributive property:
\;a\left( {x + b} \right) = c
, Solving linear equations using multiplication and division, Solving two-step linear equations:
ax + b = c
{x \over a} + b = c
How to use distributive property to solve linear equations?
11.32 = 8\left( {0.61 + x} \right)
- 5\left( {2.49 + x} \right) = 17.46
2\left( {x - 18.5} \right) = - 4.67
9.5 = - 11.5\left( {x - 5.2} \right)
x
\frac{{3.57 + x}}{2} = 2.16
- 6.79 = \frac{{x - 3.43}}{4}
- 0.761 = \frac{{0.158 + x}}{{ - 2}}
\frac{{x - 20.91}}{{ - 5}} = 8.25
A grocery store has chocolate bars on sale. Each chocolate bar is $0.25 cheaper when 4 of them are purchased. Chris bought 4 chocolate bars, and he paid $8.52. What is the original price of each chocolate bar?
ax + b = c
{x \over a} + b = c
a(x + b) = c
a(x + b) = c
ax + b = c
{x \over a} + b = c
a(x + b) = c
|
Pre-Algebra Tutor, Help and Practice Online | StudyPug
Pre-Algebra made completely easy!
Get better math mark with our complete Pre Algebra help. We cover all topics you will find in any typical Pre Algebra class, Florida FSA, S. Carolina Ready Math test and Massachusetts MCAS!
Keeping with your textbook and class, our Pre Algebra video lessons walk you through all topics such as, Polynomials, Solving multi-step equations, Fractions, Pythagorean theorem, Volume of a cylinder, Similar figures and so many more. Learn the concepts with our tutorials that show you step-by-step solutions to even the hardest pre algebra problems. Then, reinforce your understanding with tons of pre algebra practice.
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Certainly! We have lessons and practice on all topics in your pre algebra book. Other pre algebra textbooks like Prentice Hall mathematics pre algebra, McDougal Littell pre algebra, Holt pre algebra and McGraw Hill pre algebra textbook, we have got you all covered too!
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It is never a bad idea to browse around our site. However, if you want to study more efficiently, you can look at your previous pre algebra assignments and tests to identify your weaknesses. Once you know where you need help, you can watch our video lessons and try doing the pre algebra problems available on our site. If there are any problems that you can't solve, the step by step solutions from our experienced teachers can help!
What are the prerequisites for Pre Algebra?
A prerequisite for this course is 6th grade math or 7th grade math, and after you mastered Pre Algebra, your follow up course should be Algebra 1.
1Integers
1.1Understand integer multiplication
7Applying mathematical reasoning
7.2Representing patterns in linear relations
7.3Reading linear relation graphs
7.4Solving linear systems by graphing
9Exponents and Scientific Notation
11Surface Area of 3-Dimensional Shapes
12Volume of 3-Dimensional Shapes
13Function and Linear Relationships
ax = b
\frac{x}{a} = b
14.2Solving linear equations using addition and subtraction:
ax + b = c
14.3Solving linear equations using multiplication and division:
\frac{x}{a} + b = c
\;a\left( {x + b} \right) = c
d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
M = ( \frac{x_1+x_2}2 ,\frac{y_1+y_2}2)
15.3Slope intercept form: y = mx + b (free lessons)
17Symmetry and Tessellations
Pre-algebra is the course of study that introduces you to the following number types:
\bullet
\bullet
\bullet
\bullet
It also explores the basics of equations, simple roots and powers, variable manipulation and more.
If you're looking for revision aides and exam prep resources, or just need help with something specific, our easy to understand Pre-algebra lessons are here for you! Designed to ensure that all leaners can take something away from each session, our content is adaptable to your learning style and skill level. Regardless of whether you're new to the subject matter or familiar with elements of algebra, you'll find something of value in all our revision materials.
Our aim is to deliver you the best in pre-algebra help, providing you with access to helpful online videos, mock exam materials, and valuable research tools. In short, we want to make StudyPug your all-in-one pre-algebra tutor.
What Grade Is Pre-Algebra Taught?
What Is in Pre-Algebra?
\bullet
\bullet
\bullet
Surface Area and Volume of 3D Shapes
\bullet
Linear Equations and Mathematical Reasoning
\bullet
\bullet
Symmetry and Tessellation
\bullet
Representing Data
We offer content for all areas of pre-algebra (not just the ones listed here) and we provide free lessons for non-subscribers covering a variety of pre-algebra content. We greatly encourage you to explore these free lessons to get a better sense of how StudyPug can work for you. For more information, please visit https://www.studypug.com/pre-algebra.
If you're a parent looking to provide your child with additional support, we have a wealth of content that can help them tackle math in school (6th and above). We also have handy tips for helping you to prepare your middle schooler for college.
Algebra can seem intimidating at first, which is why it's necessary to develop a firm understand of the basics. To help you learn pre-algebra, StudyPug has built an extensive pre-algebra tutorial library that covers practically every question you would expect to have.
At StudyPug, we understand that every learner has their own unique way of learning and to that end, we opted to build our content with a "pre-algebra for dummies" approach that starts from the ground up, assumes no prior knowledge, and covers every aspect of pre-algebra. Each section of our content has also been designed to flow naturally from one to the next so that you can build off what you've just learned, introducing more complex elements as you go.
Subscribers to our pre-algebra content, will have access to every lesson on our website, so If you're looking for a specific element of pre-algebra, we've got you covered. Our service is also a fraction of the cost of what it would cost to hire a one-on-one math tutor, adding value for money to our fantastic study platform.
Furthermore, you have the ability to access the specific content that you need, there's no need to cover topics you already know! You can customize your sessions with the content that you feel you need to study most.
Now we're well aware that school can be a drag sometimes, you can't always be expected to be in the mood to learn when it's time for your pre-algebra math class. That's why we've made our video tutorials available to you around the clock. When you're focused and ready to learn, StudyPug is there for you.
With 1000s of lessons to observe, pre-algebra practice exams to try, and access to 24/7 help and guidance, StudyPug will make understanding pre-algebra a breeze!
For students looking to attend college upon graduating from high school, you'll find that math is a prerequisite for most colleges and this will include an understanding of basic math and the core concepts of algebra.
Take advantage of the resources at your disposal and make the most out of the time that you have. Setting aside an hour or two each evening to watch our video tutorials will make the world of difference when it comes time to take your exams. You;ll be better prepared for your classes, your tests, and your college applications.
How to Pass Pre-Algebra?
The key to passing Pre-algebra is in preparation and revision. You should make the most of our study tools and address any areas of weakness you think you may have. A good way to highlight these areas of improvement is to look at your past exams and circle the questions you've struggled with.
Our team of experts have worked to ensure that the content we provide matches the topics found in modern per-algebra textbooks. For example, you'll find that our website covers all the content found in the following textbooks:
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Prealgebra by OpenStax
\bullet
Pearson Pre-Algebra
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Glencoe Pre-Algebra
We've noticed that the language within these textbooks can sometimes be hard to digest and understand. That's why our team of mathematic tutors have broken down the key talking points of these textbooks and have expertly delivered them in an easy to understand video format that you can pause, fast-forward, and rewind.
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How to Convert the British Pound to Dollars
1 Understanding and Converting Exchange Rates
2 Estimating Currency Conversions
The official monetary currency for the United Kingdom is the British pound. The pound is also called the pound sterling. But there are a lot of currencies in the world, and you don't have to live in England to use the pound. Traveling requires you to exchange currencies, and it is not so easy as remembering that 10 pounds = 10 dollars. Currency rates change every day, even every hour, but they are still not difficult to calculate.
Understanding and Converting Exchange Rates Download Article
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Look up the current currency exchange rate, which changes by the hour. The value of a currency depends on the stock exchange rate at any given time. Real-time quotes can be found on the NASDAQ, NYSE and AMEX stock exchanges or other financial websites, but the easiest way to see this is by simply Googling "Convert dollars to pounds." No matter where you find it, look up the current "dollars to pounds" exchange rate. It should look something like this:
1 GBP = 1.4635 USD (5/24/16, 2:00PM rate)
GBP Stands for Great British Pound. The symbol is "£."
USD Stands for United States Dollar. The symbol is "$."[1] X Research source
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Multiply the amount in pounds by the conversion rate to convert it to dollars. This is all you have to do to convert currencies. Think of it this way -- if
{\displaystyle 1GBP=1.4USD}
{\displaystyle 2GBP=2.8USD}
{\displaystyle 1.4*2.}
So, for every pound, you end up with an extra 1.4 USD. So just multiply the amount in pounds by the conversion rate to get your number of dollars. Say you want to buy a £35 sweater in London's Heathrow Airport. In dollars, you'd be spending:
Conversion rate: 1 GBP = 1.4635 USD
£35 = ______ USD.
Use an online currency calculator to convert currency perfectly at any hour. Remember that currency rates can change by the minute, meaning your hand conversion may not be useful shortly after you find it. This is why the quickest and easiest method to use in converting currency is one of the free online converters. Many phones offer apps as well that are capable of keeping up with changing rates.
Open your web browser. Clear the address line and type "currency converter." Several websites for currency conversion will appear.
Enter or click on British pound in the "from" column and enter or click on USD or dollar in the "to" field.
Enter the amount of currency you want to convert from pounds to dollars.
Click "convert" to get a perfect representation.[2] X Research source
Make sure the online calculator or converter is current with up-to-date exchange rates. There should be a date on the website to indicate the last update. If this is not within 24 hours of your current date, then don't use the site unless you only need rough estimates. That said, sites like the NYSE will have to-the-minute rates, and should be trusted before any 3rd party sites.
Convert back to pounds by simply dividing the exchange rate instead of multiplying it. If you want to change pounds back to dollars, simply divide by the exchange rate. Make sure, however, that you're using the GBP → USD conversion rate, and not the USD → GBP rate. The former number tells you how many dollars one pound is worth, the later tells you how many pounds one dollar is worth. Think of it as dollars per pound vs. pounds per dollar -- they are based on the same information, but they are different numbers!
If you have dollars and want to convert to pounds, you can multiply the dollars by the USD → GBP conversion rate, OR divide it by the GBP → USD conversion rate.
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Know that currency rates fluctuate based on global demand for that currency. Currency prices are always in flux. This is because currency, like anything else, is a tradable good. Image a firm in London wants to sell to the US stores. In order to buy their goods, the US store must purchase the goods with pounds, then sell them for US dollar. This means that pounds are flowing out of England and being converted to other money, making the pounds left in England a more valuable good that the US has demand for. There are fewer pounds now, so the value of each pound goes up. Meanwhile, the US market, which spent pounds to get dollars, sees the USD decrease in value, since more of the wealth is in Dollars, giving them less demand. A high demand for a currency usually means that currency increases value.[3] X Research source Other things that affect currency include:
Concerns about the safety/health of the country's market (which is why a war-torn country has a low-value currency).
Amount of business interest or growth in a market.
Tourism.[4] X Research source
Estimating Currency Conversions Download Article
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Look up the current exchange rate to as many decimals as you can. The more decimals in your rate, the more precise your conversion will be. However, if you just need to make an estimation, you can use much simpler rates, like 1 GBP = 1.45 USD, to get a good idea of the conversion.
Multiply the amount in pounds by the whole number in the conversion rate. For now, ignore the number after the decimal, and instead multiply the pounds by the whole number. So, if you're buying a £20 hat, and the exchange rate is 1.456042, start by multiplying 20 * 1. This is almost always 1 or 0 since the USD and GBP don't often vary by that much.
Problem: Estimate Conversion of £20 into USD using an exchange rate of 1.456042
Step 1: Multiply whole number of conversion rate by price in GBP.
Use rounding to quickly estimate the exchange rate while shopping. If you see a hat for £20 and want to know what it will cost you in dollars, you'll likely struggle to multiply 20 by 1.456042. But it will be much easier to multiply by 20 by 1.5, even if you know the final estimate will be a little higher than the actual answer. Note that the smaller your rounding, the more precise the answer will be.
Step 2: Round conversion rate to manageable number.
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Find 10% of the price in pounds to get a simple, easily added percentage. This is really easy to do, though it seems complex. First, note that 10% + 10% + 10% + 10% + 10% = 50%. Second, remember that .1 is the same thing as 10%, meaning .5 is the same thing as 50%. So, if you need to find a tricky conversion decimal, simply find 10% and add it together until you reach the right percentage. And 10% is easy to find -- simply move the decimal one place to the left!
Step 3: Find 10% of the price in pounds and add it until it matches your decimal.
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Add the amount found in the decimal to the amount found in the whole number to finish your estimate. Since the whole number is usually 1, this means you simply find 10%, add as many as needed to match the decimal in the conversion rate, and add this to the amount in pounds. So, to finish the example problem:
Step 4: Add together your whole number and your decimal conversions:
Whole: £20
Decimal: £10
Estimated Amount: £30 The actual conversion (using this rate) would be $29.12, meaning this was a useful estimation.
If the exchange rate is 1 USD = 0.7221 GBP, how do I figure GBP to USD rate using the same USD to GBP rate?
Use the reciprocal of the dollar-to-pound rate. That means in this case divide 1 by 0.7221.
How do I convert dollars to British pounds?
Use the current exchange rate and proceed as shown in Method 1, Step 5 above.
How do I convert dollars to pounds?
A recent conversion factor is 0.8 pound to a dollar.
The exchange rate varies constantly. Look up the current exchange rate. It will be somewhere around one dollar equaling about two-thirds of a pound. So in this case you would multiply the dollar amount by two-thirds.
How much is $3,264 US in UK pounds?
At a recent exchange rate (1 US dollar = 0.77 British pound), 3,264 dollars equaled 2,513.28 pounds.
How do I convert rand into dollars?
At a recent exchange rate, you would multiply the number of rand by 0.072.
How much is E39 in US dollars?
At a recent exchange rate (1 euro = $1.13) 39 euros were worth $44.07.
Ask about fees. Banks often charge a fee to exchange or convert currency.
Beware of possible scams when using a private broker or foreign exchange brokerage houses in foreign countries.
↑ http://www.exchangerates.org.uk/Pounds-to-Dollars-currency-conversion-page.html
↑ http://coinmill.com/GBP_USD.html
↑ http://www.investopedia.com/ask/answers/08/how-often-to-exchange-rates-fluctuate.asp
To convert the British pound to dollars, first go online to one of the stock exchange websites or another financial website to get the current currency exchange rate. Then, multiply the amount in pounds by the exchange rate to get the value in dollars. For example, if the exchange rate is 1 British pound to 1.4 dollars and you want to buy something for 35 pounds, you can multiply 35 times 1.4 to see that you’d be paying the equivalent of 49 dollars for it. Alternatively, just search for an online currency converter and let the internet do the math for you! For tips on how to estimate currency conversions using some shortcuts when you’re on the go, keep reading!
Español:convertir libras esterlinas a dólares americanos
Français:convertir la livre sterling en euro
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using venn diagram , show that (A-B) , (A intersect B) and (B-A) are disjoint sets , taking A={2,4,6,8,10,12} and - Maths - Sets - 8143747 | Meritnation.com
using venn diagram , show that (A-B) , (A intersect B) and (B-A) are disjoint sets , taking A={2,4,6,8,10,12} and B ={3,6,9,12,15}
Given,\quad A=\left\{2,4,6,8,10,12\right\}\phantom{\rule{0ex}{0ex}}And,\quad B=\left\{3,6,9,12,15\right\}\phantom{\rule{0ex}{0ex}}So,\phantom{\rule{0ex}{0ex}}A-B=\left\{2,4,8,10\right\}\phantom{\rule{0ex}{0ex}}B-A=\left\{3,9,15\right\}\phantom{\rule{0ex}{0ex}}A\cap B=\left\{6,12\right\}\phantom{\rule{0ex}{0ex}}Now,\quad draw\quad a\quad venn\quad diagram,\quad we\quad have,
From diagram it is clear that all the three sets are disjoint.
when (a-b)= {2,4,8,10}
And (b-a)={3,9,15}
and (a intersection b)= {6,12}
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Effect of Temperature Nonuniformity on Heat Transfer in an Unshrouded Transonic HP Turbine: An Experimental and Computational Investigation | GT | ASME Digital Collection
Andy D. Smith,
QinetiQ Ltd., Farnborough, UK
Qureshi, I, Smith, AD, Chana, KS, & Povey, T. "Effect of Temperature Nonuniformity on Heat Transfer in an Unshrouded Transonic HP Turbine: An Experimental and Computational Investigation." Proceedings of the ASME Turbo Expo 2010: Power for Land, Sea, and Air. Volume 4: Heat Transfer, Parts A and B. Glasgow, UK. June 14–18, 2010. pp. 255-270. ASME. https://doi.org/10.1115/GT2010-22700
Detailed experimental measurements have been performed to understand the effects of turbine inlet temperature distortion (hot-streaks) on the heat transfer and aerodynamic characteristics of a full-scale unshrouded high pressure turbine stage at flow conditions that are representative of those found in a modern gas turbine engine. To investigate hot-streak migration, the experimental measurements are complemented by three-dimensional steady and unsteady CFD simulations of the turbine stage. This paper presents the time-averaged measurements and computational predictions of rotor blade surface and rotor casing heat transfer. Experimental measurements obtained with and without inlet temperature distortion are compared. Time-mean experimental measurements of rotor casing static pressure are also presented. CFD simulations have been conducted using the Rolls-Royce code Hydra, and are compared to the experimental results. The test turbine was the unshrouded MT1 turbine, installed in the Turbine Test Facility (previously called Isentropic Light Piston Facility) at QinetiQ, Farnborough UK. This is a short duration transonic facility, which simulates engine representative M, Re, Tu, N/
T
and Tg /Tw at the turbine inlet. The facility has recently been upgraded to incorporate an advanced second-generation temperature distortion generator, capable of simulating well-defined, aggressive temperature distortion both in the radial and circumferential directions, at the turbine inlet.
Hot-streak, OTDF, RTDF, HP turbine, Heat transfer, Aerothermodynamics, Transonic, HP rotor, Rotor casing
Heat transfer, Rotors, Temperature, Turbines
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What is matrix notation in math? | StudyPug
Notation of matrices - Linear Equations with Matrices
One of the most important tools used throughout linear algebra, and thus one of the key points to learn on this course, is matrix mathematics. Due to linear algebra being all about finding the solutions to systems of linear equations, matrix math and the study of vector spaces become a tool to represent and orderly solve such systems in an orderly and intuitive fashion.
The lesson of today focuses on the introduction of this new kind of mathematical information arrangement: the matrix (plural: matrices). On our lesson we will introduce the concept of a matrix, how can it be operated on mathematically, and then there will be a little introduction preparing you for the next lessons on this course, where you will learn how useful matrices are and the methods to solve mathematical equations with them.
In mathematics, a matrix is an ordered list of numbers put in a rectangular bracket. This list can also be called a rectangular array, and the reason for it, is that its elements can represent different coefficient terms or coefficients of variables depending on their location within the bracket box.
A matrix provides an orderly fashion to display an array of information. For example:
Equation 1: Example of a matrix
Notice the rectangular brackets are NOT just simple straight lines surrounding the array of numbers, it is important to note this since there is a different operation which can be obtain from a matrix called the determinant, which happens to have a notation very similar to the matrix itself, but instead of rectangular brackets it has straight lines around. Just remember, if the numbers are inside a "box" is a bracket matrix, if the numbers are inside two long straight lines then is a determinant. We will talk about the determinant of a 2x2 matrix and of the 3x3 matrix, along with the properties of determinants in lessons much later on this course. For now, we will concentrate on the basics of matrix notations to continue on their proper operations.
Before we learn the matrix definition let us talk about the history of this word and how it makes sense to have such a notation among math: It is no wonder how the term "matrix" in itself has been cleverly used throughout history not only among popular culture with different connotations depending on the epoch, but also through different academic areas of study from mathematics to different branches of science.
The word "matrix", from its latin etymology, means "womb" and it is used not only in math but also in areas such as biology, where is the term used to name the tissue on living organisms cells where more specialized materials (those structures with higher or more specific functions) are lodged. You can also found the term matrix being used in chemical analysis and experimentation, where the word describes all of the rest of the content in a sample which is not the particular material of interest. In geology, a matrix refers to the compact material without a characteristic shape in which well defined specimens of rock formation or crystals are embedded.
If we pay attention to all of these usages of the word throughout areas of natural science, we can find a pattern where in general we define matrix as an environment in which information of interest (be it structures or systems) is lodged, fixed or contained, and ready for us to study or use it. The same goes for matrix algebra, in mathematics we use matrix notation to organize systems of values coming from different geometrical planes. This will become obvious as we continue with our linear algebra lessons, and so we will just be mentioning this for now so you can start having an idea of what matrices are all about. On the next lessons where you will be solving systems of linear equations by graphing, and then learning how to represent a linear system as a matrix, you will be able to directly observe what we have been talking about on the usefulness and meaning of matrices as "boxes" of information.
The dimensions of a matrix define its size and it basically refers to the number of rows and columns inside the matrix. For example, if the matrix has m rows and n columns, then we say that the dimensions of a matrix are m by n.
Figure 1: Dimensions of a matrix explained
In this way, each entry in a matrix gets called a matrix element which we can denote with subindexes. For example, if we have a m by n matrix A. Then we say that the matrix element
a_{3,2}
is the entry in the third row and second column.
Equation 2: Matrix elements
Let us work through a a few examples next:
Determine the dimensions of the following matrices:
Equation 3: Examples of matrices
And so the dimensions for each matrix are:
3 x 3 = 3 rows and 3 columns
1 x 3 = 1 row and 3 columns
4 x 1 = 4 rows and 1 column
Find the matrix elements (or values) asked below from the following matrix:
Equation 4: 4 x 9 Matrix
a_{1,1}
a_{2,3}
a_{4,6}
a_{4,9}
\frac{a_{2,5}}{a_{4,9}} = \frac{7}{99}
Value of 2
(a_{3,7})
= 2(-20) = -40
(a_{2,4})^{2} = (6)^{2}
(a_{4,2})
+ 5 = 2(25) + 5 = 55
\frac{a_{1,9}}{3} = \frac{15}{3}
a_{3,5} + a_{1,1}
After learning how the dimension of a matrix provides information on the characteristics of the array notation and knowing what the elements in a matrix are, it is important to recognize there are a few different types of matrices:
Square matrices:
When a matrix has the same amount of columns and rows, we call this matrix a square matrix. A square matrix then, is a mxn matrix in which m=n, and thus, its box notation does look like a square box since the elements inside the matrix bracket are the same amount on each side. Since in this case m equals n, we usually refer to square matrices dimensions simply as nxn. Equation 2 shows the notation of a square matrix with dimensions 3x3.
Diagonal matrices:
Although diagonal matrices can refer to matrices with any combination of dimensions, the term is more often used when talking about specific types of square matrices. The reason for this is that a diagonal matrix will be easily identified by the difference between the type of elements it has in its main diagonal, versus the rest of the array.
The main diagonal of a matrix is the diagonal array of numbers which start from its uppermost left corner progressing downwards towards the right side until you get to the lowest possible row. For a square matrix, the main diagonal goes from the upper left corner to the bottom right corner. If we take equation 2 as our provided matrix, the main diagonal of this matrix contains the elements:
a_{1,1}, a_{2,2}
a_{3,3}
. For all matrices, the main diagonal is the array of elements with subindexes where m equals n, and so, the main diagonal of any matrix is the array of elements such that:
a_{1,1}, a_{2,2}, a_{3,3}... a_{n,n}
where m = n.
And so, a diagonal matrix is that in which only the elements inside its main diagonal are different to zero.
Identity matrices are square matrices by definition which contain only values of 1 in their main diagonal. Identity matrices are denoted as
I_{n}
where n represents the dimension of the matrix nxn. Such matrices are important in vector matrix notation since they provide a way to clearly discern between the terms of different variables in collections of vectors written as matrix, although we will continue with this topic in a deeper manner in later lessons, it is important for you to know that the identity matrix notation provides a base for unit vector notation in linear algebra, and thus, an identity matrix can be called a unit matrix too.
All identity matrices are diagonal matrices, and any multiples of them will be too because the only values affected by a scalar multiplication are those on the main diagonal (since the rest of them are zeros, any number multiplied by zero equals to zero, and so the zeros remain unchanged).
A zero matrix is that which has only zeros as its elements. Zero matrix notation is probably the simplest, since it does not matter the dimensions of the matrix in question, as long as all of the elements inside it are zeros, you have a zero matrix. Zero matrices may be known as null matrices too.
All of these will be studied in depth throughout the lessons of this course since they will be most useful through different operations or methods while solving linear systems of equations.
Identify which type of matrix is each of the ones in the next list, write down their dimensions and the characteristics that allowed you to identify it:
Equation 5: Different types of matrices
And so to identify each of the matrices above:
This is a 3 x 2 matrix, which means that it has three rows and two columns as it can be easily observed.
This matrix is a diagonal matrix, since it has a different amount of rows and columns, we refer to these types of matrices as "rectangular diagonal" matrices. The dimensions of matrix b) are 2 x 4.
This is a square matrix of dimensions 4 x 4 (4 rows and 4 columns).
Matrix d) is a zero matrix since all of its entries are zeros.
Matrix e) is a column matrix with dimensions 4 x 1 (all column matrices consist of one column and are also known as vector matrices).
This matrix is an identity matrix
I_{3}
which means that the dimension of matrix f) is 3, or 3 x 3 (3 rows and 3 columns).
Before we finalize this lesson with an introduction on the rules that apply to basic matrix operations we would like to recommend you to visit these notes on matrix notation and operations if you want to continue looking at examples of what we have learned so far on this course. Another nice and straightforward article you can visit is this one on matrix notation, where you can find a few more examples in a simplistic form.
For this section we will have a brief introduction to the matrix rules for operations such as addition and multiplication. These types of operations will be covered in depth and with examples later throughout this Linear Algebra course, but it is important to have a basic idea on how the matrix size and how many rows vs columns a matrix has provides guidelines on what operations can be done with such matrix or not.
The addition (and thus subtraction) of two matrices is a very simple operation to compute. The rule is that as long as you have two matrices with the same amount of rows and columns, you can add (or subtract) them. The resultant matrix will have the same amount of rows and columns as the two original matrices and the operation is simply done by adding the elements with the same subindexes from each matrix and positioning the result in the element with the same subindex in the resultant matrix. Take a look at figure 2 where you can observe the general rule for adding matrices:
Figure 2: Matrix addition
Figure 3: Rule for matrix multiplication
In order to multiply two matrices, the first matrix must have the same amount of columns as the second matrix has rows. The product of this multiplication, will be a new matrix with dimensions equal to the amount of rows in the first matrix by the amount of columns found in the second matrix. And so, this means that for two matrices
A = m_{1} \times n_{1}
B = m_{2} \times n_{2}
to be multiplied,
n_{1}
m_{2}
and the resulting matrix will have the dimensions of
C = m_{1} \times n_{2}
. This can be observed in the following figure:
Figure 4: Matrix Multiplication explained
And so, as you may have already noticed, matrix multiplication is NOT commutative: you cannot change the order of the matrices being multiplied and obtain the same result, actually, you won't be able to multiply them due to the fact that, unless they are square matrices, the first matrix will not contain the same amount of columns as the amounts of rows in the second one, making the multiplication not possible.
This is all for our topic of today, as you can see, this lesson serves as an introduction to the algebraic notation we will be using throughout most of this course and any course on Linear Algebra. Make sure you understand the examples so you can be ready to the next lesson.
Linear Equations with Matrices Topics:
R^n
R^n
In this section, we will be learning about the dimensions of a matrix, as well as finding the matrix element within a matrix. Dimensions of a matrix are determined by the number of rows and columns in the matrix. We generally write the dimensions to be (# of rows) x (# of columns). A matrix element is an entry in the matrix. If you see the subscript of the letter, then that subscript will tell you which entry we are talking about in the matrix. For example, if the subscript is (4,6), then we are looking for the entry in the 4th row and 6th column.
A matrix is a list of numbers put in a rectangular bracket.
The dimensions of a matrix are the number of rows and columns of the matrix. For example, if the matrix has m rows and n columns, then we say that the dimensions matrix is m by n.
Each entry in the matrix is called a matrix element. Let the matrix be called A. Then we say that the matrix element
a_{4,6}
is the entry in the 4th row and 6th column.
Notation of Matrices Overview
The Dimensions of a Matrix
Find the matrix element (or value) of the following matrix
a_{1,1}
a_{2,3}
a_{4,6}
a_{4,9}
\frac{a_{2,5}}{a_{4,9}}
2(a_{3,7})
(a_{2,4})^2
2(a_{4,2})+5
\frac{a_{1,9}}{3}
a_{3,5}+a_{1,1}
We have over 70 practice questions in Linear Algebra for you to master.
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Applying the Pythagorean Theorem | StudyPug
We will need to apply Pythagorean Theorem often in our daily life. In this lesson, we will focus on tackling some Pythagorean Theorem word problems.
Basic Concepts: Pythagorean theorem, Estimating square roots, Using the pythagorean relationship
d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
An aeroplane flew from point A to point B, and then to Point C. How far is the plane from point A? Refer to the diagram below.
A 2.5 m long ladder is put against a wall. If the base of the ladder is 1.25 m away from the wall, what is the length from the ground to the point where the top of ladder touches the wall?
Which triangle has a longer perimeter?
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ηmax=K122Q1Q2/(1+1+K122Q1Q2)2
Pout−max=K12Q1Q2R1R2RLV2(R1(R2+RL)+K12Q1Q2R1R2)2
K12=M12/L1L2
ST(u,v)|u,v∈(0,1)
SR(u,v)|u,v∈(0,1)
P01=P0∪P1
argmin‖n‖=1∑i=0m3((P01(i)−c)Tn)2
c=∑i=0m3P01(i)/m3
PC0={(P0(i)−c)−((P0(i)−c)⋅nT)⋅n|i=0…m1}
PC1={(P1(i)−c)−((P1(i)−c)⋅nT)⋅n|i=0…m2}
Sprial(θ)={(1−θ2π)Cpc0∩V(θ)+θ2πCpc1∩V(θ)|θ=0…2π}
Rt/r=ρ∫u=01ct/r(u)du/(WH)
M=μ04π∮Ct∮Crdct→*dcr→D
Lt/r=(Mmid−inn+Mmid−out)2+μ08π∫01ct/r(u)′du
Pout−max=4π2f2M(a,b)LT(a,b)3/2LR(a,b)3/2RLV216π4f4M(a,b)2LT(a,b)3LR(a,b)3+2(Rt(a,b)Rr(a,b)+Rt(a,b)RL+1)4π2f2M(a,b)LT(a,b)3/2LR(a,b)3/2+(Rt(a,b)Rr(a,b)+Rt(a,b)RL)2
ηmax=4π2f2M(a,b)2Rt(a,b)Rr(a,b)+(Rt(a,b)2Rr(a,b)2+4π2f2M(a,b)2Rt(a,b)Rr(a,b))1/2
argmaxa,b∈(0,1)fefficiency(a,b)
SRk(a,b)
SRk+1(a,b)
STk(a,b)
STk+1(a,b)
Finite Element Analysis (FEA) for Optimization the Design of a Baja SAE Chassis
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Measurement of the Dynamic Shear Modulus of Mouse Brain Tissue In Vivo by Magnetic Resonance Elastography | J. Biomech Eng. | ASME Digital Collection
Stefan M. Atay,
Stefan M. Atay
, 1 Brookings Drive, Box 1185, St. Louis, MO 63130
Christopher D. Kroenke,
Advanced Imaging Research Center,
, 3181 S. W. Sam Jackson Park Road, Portland, OR 97239-3098
Arash Sabet,
Department of Mechanical and Aerospace Engineering, Department of Biomedical Engineering,
e-mail: pvb@me.wustl.edu
Atay, S. M., Kroenke, C. D., Sabet, A., and Bayly, P. V. (March 31, 2008). "Measurement of the Dynamic Shear Modulus of Mouse Brain Tissue In Vivo by Magnetic Resonance Elastography." ASME. J Biomech Eng. April 2008; 130(2): 021013. https://doi.org/10.1115/1.2899575
In this study, the magnetic resonance (MR) elastography technique was used to estimate the dynamic shear modulus of mouse brain tissue in vivo. The technique allows visualization and measurement of mechanical shear waves excited by lateral vibration of the skull. Quantitative measurements of displacement in three dimensions during vibration at
1200Hz
were obtained by applying oscillatory magnetic field gradients at the same frequency during a MR imaging sequence. Contrast in the resulting phase images of the mouse brain is proportional to displacement. To obtain estimates of shear modulus, measured displacement fields were fitted to the shear wave equation. Validation of the procedure was performed on gel characterized by independent rheometry tests and on data from finite element simulations. Brain tissue is, in reality, viscoelastic and nonlinear. The current estimates of dynamic shear modulus are strictly relevant only to small oscillations at a specific frequency, but these estimates may be obtained at high frequencies (and thus high deformation rates), noninvasively throughout the brain. These data complement measurements of nonlinear viscoelastic properties obtained by others at slower rates, either ex vivo or invasively.
biological tissues, biomechanics, biomedical MRI, brain, deformation, elastic waves, shear modulus, vibrations, viscoelasticity, elastography, MRI, brain, stiffness
Biological tissues, Brain, Elastography, Shear modulus, Displacement, Simulation, Vibration, Imaging
A Quantitative Comparison of Modulus Images Obtained Using Nanoindentation with Strain Elastograms
Correspondence of Ultrasound Elasticity Imaging to Direct Mechanical Measurement in Aging DVT Rats
Estimation of Displacement Vectors and Strain Tensors in Elastography Using Angular Isonifications
In Vivo Measurement of Rapid Deformation and Strain in an Animal Model of Traumatic Brain Injury
Physical Model Simulations of Brain Injury in a Primate
Are In Vivo and In Situ Brain Tissues Mechanically Similar
Microscopic Magnetic Resonance Elastography (μMRE)
Evaluation of a Material Parameter Extraction Algorithm Using MRI-Based Displacement Measurements
Mechanical Transient-Based Magnetic Resonance Elastography
Determination and Analysis of Guided Wave Propagation Using Magnetic Resonance Elastography
Starkoski
C. L. J.
Magnetic Resonance Imaging of Ultrasound Fields: Gradient Characteristics
Validation of a Dynamic MR Elastography Technique Customized for In Vitro Biomechanical Assessment of Articular Cartilage Under High Frequency Cyclical Shear
The Large Shear Strain Dynamic Behaviour of In-Vitro Porcine Brain Tissue and a Silicone Gel Model Material
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How do you find the area of a sector of a circle? | StudyPug
To begin this lesson on finding the area of a sector, we'll first have to start with the area of a circle. A circle with a radius r has an area A found through:
A=\pi r^2
Do you know how to then find the circumference? The circumference C can be found with the formula of:
C=2 \pi r
As you can see, in both cases, for both the area and the circumference, we're dealing with the whole circle. Let's take a look at if we were just dealing with a section of the circle.
We can divide a circle into sectors. You can imagine sectors as a radius that extends from the circle's center that then goes around the circumference for a bit, then turns to go down a radius once again to meet at the center. You'll get a sector of a circle that looks like a slice of pie. It has an angle at the center of the slice (or the tip of the slice, if you prefer) that is called the "central" angle.
An important term we're going to have to learn is arc length. The arc length of a sector is the portion of circumference that the sector takes up from the whole circle. It is the portion of the circumference subtended by the central angle.
Keeping in mind the circle's total area, the circumference, and the arc length, we can now learn how to find the area of a section of a circle.
Through an example question, we'll demonstrate how the relationship between the area of a section and total area can help you find the section's area. There is an area of a sector formula that can help you out which we'll share below, so don't worry about it for now.
Calculate the area of sectors of circles given arc length and radius
(i) Area of the circle
(ii) The area of the shaded sector
(i) To find area of a circle, use the formula
\pi r^2
\pi r(20)^2=1256.6m
We simply need to substitute in the radius of 20m to find the answer.
\frac{area\;of\;section}{total\;area} = \frac{arc\;length}{circumference}
\frac{area\;of\;section}{1256.6} = \frac{279.25}{2 \pi (20)}
area\;of\;section = \frac{279.25(1256.6)}{2 \pi (20)}
area\;of\;section = 2792\;m^2
What exactly are we doing here? In this question, we're finally trying to find the area of a sector. Since the sector is a portion of the total circle, there's a relationship between them. Likewise, there's a relationship between the section of the circumference that's a part of the sector (the arc length), versus the whole circumference. Their ratios are the same! Or rather, their proportions are the same.
That's why we use the formula of:
\frac{area\;of\;section}{total\;area} = \frac{arc\;length}{circumference}
This helps us use the proportions to work out the unknown. In this case, the unknown is the area of the sector. We're given the arc length (279.25m), we've got the circumference from the formula of
C = 2 \pi r
, and lastly, we've calculated the total area from part (i) of the question. The rest of the question is just a matter of substituting in these numbers to the formula and then isolating the area of a sector (the unknown) onto one side to find the answer.
This online interactive sector area of a circle can be dragged so you can explore how the area sector changes as its central angle changes. It's worth checking out to solidify your understanding of the relationships explored in this lesson for circles.
Find the areas of the circles with the given information below.
i) Area of the circle
ii) The area of the shaded sector
Linda wants to get pizzas for her house party. She has invited 10 friends over. A 12-inch pizza is enough for 4 people. An 8-inch pizza is $7, while a 12-inch pizza is $12. How many 8-inch pizzas and 12-inch pizzas does she need to order, if:
she wants to spend the least amount of money?
she wants to have 4 pizzas and spend as little as possible?
she wants to eat as much outer crust as possible?
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Sentiment Metrics | Santiment Academy
Ivan Klimuk, Santiment authors
The sentiment metrics are build on top of the Social Data
Sentiment Analysis is the problem of computationally identifying and categorizing emotions, opinions and subjective information in a given piece of text. This problem can be solved using different techniques: rule-based or machine learning. The first one represents a set of predefined rules that are used to estimate the sentiment of the input text. This approach is often less accurate and requires a lot of manual work. The amount of documents in our Social Data storage makes it barely impossible to analyze them manually. That's why we use machine learning to approach the sentiment analysis problem.
We trained a machine learning model on a large Twitter dataset, that contains over 1.6 million tweets, each labelled as either positive or negative. This model is then used to evaluate the sentiment of each single document in the Social Data set, i.e. it assigns a positive and negative sentiment score to each message/post/comment/etc. These scores are probabilities that the content of the text being analyzed is positive or negative respectively. Therefore both the positive and negative sentiment scores fall in a range between 0 (not positive/negative at all) and 1 (extremely positive/negative). Moreover, the sum of these two scores always equals 1.
I'm really excited about the new Libra currency!
This message has a positive score of 0.75 and a negative score of 0.25.
We use this approach for messages and comments from social networks conversations because the structure of the text there is usually more or less the same: short messages with a single and/or simple idea behind them. But this is not the case for all the messages: some of them might be long and complicated, some might be just neutral or contain spam or other irrelevant information. These kind of messages usually have a pretty vanished pair of sentiment scores: both positive and negative scores are close to 0.5. We don't include these kind of messages while calculating the Sentiment Metrics: they are filtered out by a certain threshold.
1. Positive (Negative) Sentiment
The total sum of positive (negative) sentiment scores of a given set of documents over time. Only scores that are equal or higher than 0.7 are taken into account. Can be calculated for a certain asset or for any given search term, similar to the social volume.
Relative number, less or equal than the corresponding social volume.
We store each of the social data documents with its absolute timestamp. I.e. it is possible to aggregate the data with any desired interval on request. Currently the time intervals we use are the following:
In Sanbase: Five-Minute Intervals
In Sanbase Graphs: 6h, 12h, 1d.
The sentiment scores are calculated every 5 minutes. Taking into account that the social data itself is quasi-realtime, the maximal latency is 5 minutes.
We do not separate or filter the social data being collected by assets. I.e. we can calculate this metric for any asset. More on this can be found here.
Sanbase Graphs
The metric is available for any selected asset.
2. Average Sentiment
The difference between the Positive and Negative Sentiment metrics.
Relative number. This metric falls in the range [-social_volume, +social_volume] where social_volume is the corresponding social volume.
Same as Positive (Negative) Sentiment.
3. Sentiment Weighted
The Sentiment Weighted is an improved version of the Sentiment Balance that also takes into account the Unique Social Volume.
Sentiment Weighted is defined as a rolling Z-score of
X = \mathrm{Unique Social Volume} \times \mathrm{Average Sentiment}
More precisely we choose a duration
d
which will be the length of our sliding window. Then for any timestamp
t
we consider the population
X(t,d)
consisting of all values of
X(t')
for all timestamps
t'
t-d
t
. If we use
\mu
\sigma
to denote mean and standard deviation, then we define Sentiment Weighted as:
Sentiment Weighted(t,d) = \frac{X(t) - \mu(X(t,d))}{\sigma(X(t,d))}
Intuitively this score can be explained as a social-volume-weighted sentiment balance. I.e. this metric will spike when the social volume is really high and the vast majority of the messages in it are very positive at the same time. Dips will occur when the social volume again is high, but the overall sentiment is negative. In case the volume is high but the sentiment is mixed, or the sentiment has a strong positive (negative) polarity but with a low volume, the Sentiment Weighted metric won't have significant changes and will stay around 0.
Relative number. Theoretically this metric has no lower or upper limit, but normally it lies in the range of [-3, 3]. Values from outside this range indicate that something abnormal is happening.
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Template:Cquote/doc - Seeds of the Word
{{Cquote|quote=text of the quotation|author=person quoted|source=source of the quotation}}
Parameter 1, quote
Note: if the quote text contains a = (equal sign), then the template must be called with a named parameter for the content. For example: {{Cquote|quote=quote text=more quoted text}} (see Help:Template § Usage hints and workarounds).
quotealign
set to center to center align the quote text.
Sourced example
{\displaystyle E=mc^{2}}
— First Amendment to the Constitution of the United States
Note that it will wrap around article text like this.
Sets scrript direction right-to-left (Arabic, Hebrew)
Retrieved from "https://en.seminaverbi.bibleget.io/w/index.php?title=Template:Cquote/doc&oldid=446"
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Hadrosaurs, a family of duck-billed, plant-eating dinosaurs, were large creatures with thick, strong tails. It has recently been determined that hadrosaurs probably originated in North America.
Scientists in Alaska recently found a hadrosaur footprint like the one at right that measured
14
inches across. It is believed that the footprint was created by a young dinosaur that was approximately
27
feet long. Adult hadrosaurs have been known to be
40
feet long. How wide would you expect a footprint of an adult hadrosaur to be? Show your reasoning.
Set up a proportional equation comparing foot size and length in adult and juvenile hadrosaurs.
See the example below, one of several possibilities.
\frac{\text{juvenile foot size}}{\text{adult foot size}}=\frac{\text{juvenile length}}{\text{adult length}}
20.7
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Damba Yahaya1,2* , Nicholas Denwar3, Matthew W. Blair1
Yahaya, D. , Denwar, N. and Blair, M. (2019) Effects of Moisture Deficit on the Yield of Cowpea Genotypes in the Guinea Savannah of Northern Ghana. Agricultural Sciences, 10, 577-595. doi: 10.4236/as.2019.104046.
1-\frac{\text{YDS}}{\text{YNS}}
\frac{\text{1}-\left(\frac{\text{YDSi}}{\text{YNSi}}\right)}{\text{1}-\left(\frac{\text{YDS}}{\text{YNS}}\right)}
\frac{\text{1}-\left(\frac{\text{YDS}}{\text{YNS}}\right)}{\text{DII}}
\frac{\text{YDSi}\ast \text{YNSi}}{{\left(\text{YNS}\right)}^{2}}
\frac{\text{YDSi}\ast \text{YNSi}}{\text{YNS}\ast \text{YNS}}
\frac{\text{YDSi}+\text{YNSi}}{2}
\sqrt{\text{YDSi}\times \text{YNSi}}
\frac{\text{Ydsi}}{\text{Ys}}
\left(\frac{\text{YNSi}-\text{YDSi}}{\text{YNSi}}\right)\times 100
\frac{\text{YDSi}}{\text{YNSi}}
[1] Kamara, A.Y., Omoigui, L.O., Kamai, N., Ewansiha, S.U. and Ajeigbe, H.A. (2018) Improving Cultivation of Cowpea in West Africa. In: Sivasankar, S., et al., Eds., Achieving Sustainable Cultivation of Grain Legumes, Burleigh Dodds Science Publishing Limited, Volume 2, 235-252.
[2] Inaizumi, H., Singh, B.B., Sanginga, P.C., Manyong, V.M., Adesina, A.A. and Tarawali, S. (1999) Adoption and Impact of Dry-Season Dual-Purpose Cowpea in the Semiarid Zone of Nigeria (IITA IMPAC). IITA and Meg-Comm Network, Ibadan, Nigeria.
[3] Hall, A.E. (2012) Phenotyping Cowpeas for Adaptation to Drought. Frontiers in Physiology, 3, 1-8.
[4] Egbe, O., Alibo, S. and Nwueze, I. (2010) Evaluation of Some Extra-Early and Early-Maturing Cowpea Varieties for Intercropping with Maize in Southern Guinea Savanna of Nigeria. Agriculture and Biology Journal of North America, 1, 845-858.
[5] Gomez Carlos, P. (2004) Cowpea Post-Harvest Operations. Food and Agriculture Organization of the United Nations, 1-70.
[6] Singh, B.B. (2014) Future Prospects of Cowpea. In: The Food Legume of the 21st Century, Crop Science Society of America, Madison, 145-157.
https://doi.org/10.2135/2014.cowpea.c7
[7] Tony, N., James, O. and Nixon, T. (2015) Cowpea Production Handbook. IFS/AGRA, Juba.
[8] Singh, B.B. (2014) Origin, Distribution, and Importance. In: The Food Legume of the 21st Century, Crop Science Society of America, Madison, 1-15.
[9] Timko, M.P. and Singh, B.B. (2008) Cowpea, a Multifunctional Legume. In: Moore P.H. and Ming, R., Eds., Genomics of Tropical Crop Plants, Springer, New York, 227-258.
[10] MoFA/SRID (2011) Agriculture in Ghana. Facts and Figures, Accra.
[11] Egbadzor, K.F., Yeboah, M., Offei, S.K., Ofori K. and Danquah, E.Y. (2013) Farmers’ Key Production Constraints and Traits Desired in Cowpea in Ghana. Journal of Agricultural Extension and Rural Development, 5, 14-20.
[12] MoFA (2016) Agriculture in Ghana, Facts and Figures. Ministry of Food and Agriculture, Statistics, Research and Information Directorate (SRID), 25, 1-113.
https://www.agrofood-westafrica.com/fileadmin/user_upload/messen/agrofood-Westafrica/Brochure/Agriculture-in-Ghana-Facts-and-Figures-2015.pdf
[13] Singh, B.B. (2014) Production Constraints. In: The Food Legume of the 21st Century, Crop Science Society of America, Madison, 35-53.
[14] Lesk, C., Rowhani, P. and Ramankutty, N. (2016) Influence of Extreme Weather Disasters on Global Crop Production. Nature, 529, 84-87.
[15] Fahad, S., Bajwa, A.A., Nazir, U., Anjum, S.A., Farooq, A., Zohaib, A., Sadia, S., Nasim, W., Adkins, S., Saud, S., Ihsan, M.Z., Alharby, H., Wu, C., Wang, D. and Huang, J. (2017) Crop Production under Drought and Heat Stress: Plant Responses and Management Options. Frontiers in Plant Science, 8, 1147.
[16] Barnabás, B., Jäger, K. and Fehér, A. (2008) The Effect of Drought and Heat Stress on Reproductive Processes in Cereals. Plant, Cell and Environment, 31, 11-38.
[17] Dellal, I. and McCarl, B.A. (2010) The Economic Impacts of Drought on Agriculture: The Case of Turkey. In: López-Francos, A. and López-Francos, A., Eds., Economics of Drought and Drought Preparedness in a Climate Change Context, CIHEAM/FAO/ICARDA/GDAR/CEIGRAM/MARM, Zaragoza, CIHEAM Options Méditerranéennes, No. 95, 169-174.
[18] Jaleel, C.A., Manivannan, P., Abdul Wahid, M.F., Al-Juburi, H.J., Somasundaram, R. and Panneerselvam, R. (2009) Drought Stress in Plants: A Review on Morphological Characteristics and Pigments Composition. International Journal of Agriculture and Biology, 11, 100-105.
[19] Rehman, M.U., Rather, G.H., Gull, Y., Mir, M.R. Mir, M.M., Waida, U.I. and Hakeem, K.R. (2015) Effect of Climate Change on Horticultural. In: Hakeem, K.R., Ed., Crop Production and Global Environment Issues, Springer, Berlin, 211-239.
[20] Eziz, A., Yan, Z., Tian, D., Han, W., Tang, Z. and Fang, J. (2017) Drought Effect on Plant Biomass Allocation: A Meta-Analysis. Ecology and Evolution, 7, 11002-11010.
[21] Neumann, P.M. (2008) Coping Mechanisms for Crop Plants in Drought-Prone Environments. Annals of Botany, 101, 901-907.
[22] Van Loon, A.F. (2015) Hydrological Drought Explained. WIREs Water, 2, 359-392.
[23] Farrell, D., Trotman, A. and Cox, C. (2011) Global Assessment Report on Disaster Risk Reduction: Drought Early Warning and Risk Reduction. A Case Study of the Caribbean Drought of 2009-2010. International Federation of the Red Cross.
[24] Kumar, D. and Narain, P. (2005) Production Technology for Cowpea. ACIRP on Arid Legumes, CAZRI. Jodhpur, Rajasthan.
[25] SARI (2015) Savvannah Agriculture Research Institute 2015 Metrological Department Report.
[26] Walkley, A.J. and Black, I.A. (1934) Estimation of Soil Organic Carbon by the Chromic Acid Titration Method. Soil Science, 37, 29-38.
[28] Toth, S.J. and Prince, A.L. (1949) Estimation of Cation-Exchange Capacity and Exchangeable Ca, K, and Na Contents of Soils by Flame Photometer Techniques. Soil Science, 67, 439-446.
[29] Fischer, R.A. and Maurer, R. (1978) Drought Resistance in Spring Wheat Cultivars. I Grain Yield Responses. Australian Journal of Agricultural Research, 29, 897-912.
[30] Fernandez, G.C.J. (1992) Effective Selection Criteria for Assessing Stress Tolerance. In: Kuo, C.G.E., Ed., Proceedings of the International Symposium on Adaptation of Vegetables and Other Food Crops in Temperature and Water Stress, AVRDC Publication, Tainan, 257-270.
[31] Rosielle, A.A. and Hamblin, J. (1981) Theoretical Aspects of Selection for Yield in Stress and Non-Stress Environment. Crop Science, 21, 943-946.
[32] Gavuzzi, P., Rizza, F., Palumbo, M., Campanille, R., Ricciardi, G. and Borghi, B. (1997) Evaluation of Field and Laboratory Predictors of Drought and Heat Tolerance in Winter Cereals. Canadian Journal of Plant Science, 77, 523-531.
[33] Bouslama, M. and Schapaugh Jr., W.T. (1984) Stress Tolerance in Soybeans. I. Evaluation of Three Screening Techniques for Heat and Drought Tolerance. Crop Science, 24, 933-937.
[34] DPP (2014) Production Guidelines for Cowpea. Pretoria.
[35] Davis, D.W., Oelke, E.A., Oplinger, E.S., Doll, J.D., Hanson, C.V. and Putnam, D.H. (1991) Alternative Field Crops Manual: Crambe. University of Wisconsin, Madison.
https://www.hort.purdue.edu/newcrop/afcm/crambe.html
[36] Pejić, B., Mačkić, K., Mikić, A., Ćupina, B., Peksen, E. and Antanasović, S. (2013) Effect of Water Stress on the Yield of Cowpea (Vigna unguiculata L. Walp.) in Temperate Climatic Conditions. Contemporary Agriculture/Savremena Poljoprivreda, 62, 168-176.
[37] Shao, H.B., Chu, L.Y., Jaleel, C.A. and Zhao, C.X. (2008) Water-Deficit Stress-Induced Anatomical Changes in Higher Plants. Comptes Rendus-Biologies, 331, 215-225.
[38] Ishiyaku, M.F. and Aliyu, H. (2013) Field Evaluation of Cowpea Genotypes for Drought Tolerance and Striga Resistance in the Dry Savanna of the North-West Nigeria-SciAlert Responsive Version. International Journal of Plant Breeding and Genetics, 7, 47-56.
https://doi.org/10.3923/ijpbg.2013.47.56
[39] Mwale, S.E., Ochwo-ssemakula, M., Sadik, K., Achola, E., Okul, V., Gibson, P. and Rubaihayo, P. (2017) Response of Cowpea Genotypes to Drought Stress in Uganda. American Journal of Plant Science, 8, 720-733.
[40] Turk, K.J. and Hall, A.E. (1980) Drought Adaptation of Cowpea. IV. Influence of Drought on Water Use, and Relations with Growth and Seed Yield. Agronomy Journal, 72, 434-439.
[41] Turk, K.J., Hall, A.E. and Asbel, C.W. (1980) Drought Adaptation of Cowpea (I) Influence of Drought on seed Yield. Agronomy Journal, 72, 413-420.
[42] Farooq, M., Gogoi, N., Barthakur, S., Baroowa, B., Bharadwaj, N., Alghamdi, S.S. and Siddique, K.H.M. (2016) Drought Stress in Grain Legumes during Reproduction and Grain Filling. Journal of Agronomy and Crop Science, 1-22.
[43] Ajayi, A.T., Gbadamosi, A.E. and Olumekun, V.O. (2018) Screening for Drought Tolerance in Cowpea (Vigna unguiculata L. Walp) at Seedling Stage under Screen House Condition. International Journal of Biosciences and Technology, 11, 1-19.
[44] Qasem, J.R. and Biftu, K.N. (2010) Growth Analysis and Responses of Cowpea (Vigna sinensis L. Savi Ex Hassk.) and Redroot Pigweed (Amaranthus retroflexus L.), Grown in Pure and Mixed Stands, to Density and Water Stresses. The Open Horticulture Journal, 3, 21-30.
[45] Anantharaju, P. and Muthiah, A.R. (2008) Screening for Drought Tolerance in Cowpea Vigna unguiculata (L.) walp. Legume Research, 31, 283-285.
[46] Iwuagwu, M., Ogbonnaya, C.I. and Onyike, N.B. (2017) Physiological Response of Cowpea [Vigna unguiculata (L.) Walp.] to Drought: The Osmotic Adjustment Resistance Strategy Physiological Response of Cowpea [Vigna unguiculata (L.) Walp.]. Academic Journal of Science, 7, 329-344.
[47] Islam, S., Calvin, K. and Willy, V. (2011) Physiological and Morphological Characteristics of Cowpea Genotypes to Drought Stress. Arkansas Environmental, Agricultural and Consumer Sciences Journal, 42-45.
[48] Dadson, R.B., Hashem, F.M., Javaid, I., Joshi, J., Allen, A.L. and Devine, T.E. (2005) Crop/Stress Physiology Effect of Water Stress on the Yield of Cowpea (Vigna unguiculata L. Walp.) Genotypes in the Delmarva Region of the United States. Journal of Agronomy and Crop Science, 191, 210-217.
[49] Agbicodo, E.M., Fatokun, C.A., Muranaka, S., Visser, R.G.F. and Linden Van Der, C.G. (2009) Breeding Drought Tolerant Cowpea: Constraints, Accomplishments, and Future Prospects. Euphytica, 167, 353-370.
[50] Turk, K.J. and Hall, A.E. (1980) Drought Adaptation of Cowpea. II. Influence of Drought on Plant Water Status and Relations with Seed Yield. Agronomy Journal, 72, 421-427.
[51] Anuradha, Goyal, R.K. and Bishnoi, C. (2017) Assimilate Partitioning and Distribution in Fruit Crops: A Review. Journal of Pharmacognosy and Phytochemistry, 6, 479-484.
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The sides of each of the triangles below can be found using one of the shortcuts from Section 5.2. Try to find the missing lengths using your patterns. Do not use a calculator.
Is this triangle similar to a Pythagorean Triple?
The triangle is similar to the Pythagorean Triple
\text{3-4-5}
, with a scale factor of
4
The missing side is similar to
4
\text{3-4-5}
With a scale factor of
4
, the missing side is
4·4=16
45^\circ\text{-}45^\circ\text{-}90^\circ
triangle. What is the scale factor between this
45^\circ\text{-}45^\circ\text{-}90^\circ
special triangle and the given triangle?
4
. Missing sides are...
4
4\sqrt{2}
Is this triangle similar to a Pythagorean Triple of
\text{3-4-5}
\text{5-12-13}
30^\circ\text{-}60^\circ\text{-}90^\circ
triangle, the short side is half the length of the hypotenuse so it is
10
The side opposite the
60^\circ
angle is multiplied by the square root of
3
. So it is...
10\sqrt{3} \text{ m}
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GU\left(n,q\right)
U\left(n,q\right)
n×n
{q}^{2}
q
GU\left(n,q\right)
n=2
q\le 20
n=3
q\le 5
n=4
q\le 4
n=5,6
q=2
\mathrm{with}\left(\mathrm{GroupTheory}\right):
\mathrm{GeneralUnitaryGroup}\left(2,2\right)
\textcolor[rgb]{0,0,1}{\mathbf{GU}}\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right)
\mathrm{GroupOrder}\left(\mathrm{GeneralUnitaryGroup}\left(2,4\right)\right)
\textcolor[rgb]{0,0,1}{300}
\mathrm{IdentifySmallGroup}\left(\mathrm{GeneralUnitaryGroup}\left(2,4\right)\right)
\textcolor[rgb]{0,0,1}{300}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{22}
\mathrm{GroupOrder}\left(\mathrm{GeneralUnitaryGroup}\left(4,q\right)\right)
\left(\textcolor[rgb]{0,0,1}{q}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)
\mathrm{simplify}\left(\mathrm{GroupOrder}\left(\mathrm{GeneralUnitaryGroup}\left(2,{3}^{k}\right)\right)\right)
{\textcolor[rgb]{0,0,1}{81}}^{\textcolor[rgb]{0,0,1}{k}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{27}}^{\textcolor[rgb]{0,0,1}{k}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{9}}^{\textcolor[rgb]{0,0,1}{k}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{3}}^{\textcolor[rgb]{0,0,1}{k}}
\mathrm{simplify}\left(\mathrm{ClassNumber}\left(\mathrm{GeneralUnitaryGroup}\left(2,{3}^{k}\right)\right)\right)
{\textcolor[rgb]{0,0,1}{9}}^{\textcolor[rgb]{0,0,1}{k}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{3}}^{\textcolor[rgb]{0,0,1}{k}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}
Here is a general formula for the order of the general unitary group of dimension
q
\mathrm{GroupOrder}\left(\mathrm{GeneralUnitaryGroup}\left(n,q\right)\right)
\left(\textcolor[rgb]{0,0,1}{q}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{q}}^{\frac{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)}{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\prod }_{\textcolor[rgb]{0,0,1}{k}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}}^{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{k}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{-}{\left(\textcolor[rgb]{0,0,1}{-1}\right)}^{\textcolor[rgb]{0,0,1}{k}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}\right)\right)
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Fundamental Solutions and Asymptotic Behaviour for the $p$-Laplacian Equation | EMS Press
Fundamental Solutions and Asymptotic Behaviour for the
p
-Laplacian Equation
We establish the uniqueness of fundamental solutions to the
p
\mathrm {(PLE)} \; u_t = \mathrm {div} (|Du|^{p-2}Du), \; p > 2,
x \in \mathbb R^N
0 < t < T
. We derive from this result the asymptotic behaviour of nonnegative solutions with finite mass, i.e. such that
u(\cdotp, t) \in L^1(\mathbb R^N)
. Our methods also apply to the porous medium equation
\mathrm {(PME)} \; u_t = \Delta (u^m), \; m > 1,
giving new and simpler proofs of known results. We finally introduce yet another method of proving asymptotic results based on the idea of asymptotic radial symmetry. This method can be useful in dealing with more general equations.
Shoshana Kamin, Juan Luis Vázquez, Fundamental Solutions and Asymptotic Behaviour for the
p
-Laplacian Equation. Rev. Mat. Iberoam. 4 (1988), no. 2, pp. 339–354
|
Reza Kargar, Tofigh Allahviranloo, Mohsen Rostami-Malkhalifeh, Gholam Reza Jahanshaloo, "A Proposed Method for Solving Fuzzy System of Linear Equations", The Scientific World Journal, vol. 2014, Article ID 782093, 6 pages, 2014. https://doi.org/10.1155/2014/782093
Reza Kargar ,1 Tofigh Allahviranloo ,1 Mohsen Rostami-Malkhalifeh,1 and Gholam Reza Jahanshaloo1
1Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran 18558-17556, Iran
This paper proposes a new method for solving fuzzy system of linear equations with crisp coefficients matrix and fuzzy or interval right hand side. Some conditions for the existence of a fuzzy or interval solution of linear system are derived and also a practical algorithm is introduced in detail. The method is based on linear programming problem. Finally the applicability of the proposed method is illustrated by some numerical examples.
Systems of simulations linear equations play a major role in various areas such as mathematics, physics, statistics, engineering, and social sciences. Since in many applications at least some of the system’s parameters and measurements are represented by fuzzy rather than crisp numbers, it is important to develop mathematical models and numerical procedures that would appropriately treat general fuzzy linear systems and solve them [1]. A general model for solving an fuzzy system of linear equation (FSLE) whose coefficients’ matrix is crisp and right hand side column is an arbitrary fuzzy number vector was first proposed by Friedman et al. [1]. Different authors [2–5] have investigated numerical methods for solving such FSLE. Most of mentioned methods in different articles are based on numerical methods such as matrices decomposition and iterative solutions. Previously mentioned papers do not discuss a lot the possibility of solutions. In addition they cannot find alternative solutions. But the proposed method does not include these defects. Allahviranloo et al. [6] have presented that the above-mentioned method is not applicable and does not have solution generally. This paper sets out to investigate the solution of the fuzzy linear system using a linear programming method. Also we are going to explain the necessary and sufficient conditions for existence of the solutions. The idea of this method can join some uses of linear programming to solve the problems of interval data in [7–9]. The structure of this paper is organized as follows.
In Section 2, we provide some basic definitions and results which will be used later.
In Section 3, we prove some theorems which are used for proposed method and present a practical procedure. The introduced method is illustrated by solving some examples in Section 4 and conclusions are drawn in Section 5.
In this section some basic definitions and concepts are brought.
Definition 1 (by [10]). The triangular fuzzy numbers (TFN) are very popular and are denoted by and defined by Note. If , then TFN is defined: if , then TFN is defined: and finally if , then TFN is defined:
Definition 2 (by [10]). Let be a triangular fuzzy number; then one defines
Lemma 3 (by [10]). For arbitrary interval , the following properties hold:(i),(ii),(iii)for each ,
Definition 4 (by [5]). Let be a set of all fuzzy numbers on and a set of intervals on . The linear system of equations is as follows: where the coefficient matrix , , , is a crisp matrix and , , is called a FSLE.
Theorem 5. If and are triangular fuzzy numbers and if , then (i) if and only if and ,(ii)(iii)
Proof. [10].
Theorem 6. Let be matrix and an m-vector. The system with condition has a solution if and only if the optimal solution of the below system is zero:
First of all the following definitions and theorems are introduced.
Definition 7. Let ; it can be written as such that Now one can have the following theorems.
Theorem 8. Let and be intervals and ; then(i) if and only if and ,(ii)if , then and ,(iii)let such that ; then and .
Proof. Proofs of parts (i) and (ii) are obvious so we prove part (iii). Let such that
where for the proof is the same.
Theorem 9. If is a matrix and is an interval vector, then where and and .
Proof. According to Theorem 8, for , we have
Theorem 10. If is a matrix and , are two interval vectors, then the system has solution(s) if and only if the following systems have solution:
Proof. Let be an interval solution of ; then according to Definition 7 with and according to Theorem 9, we have but and by using part (i) of Theorem 8 this means the converse holds obviously.
Theorem 11. The system with condition has a solution if and only if optimized value of the below linear programing is zero:
Proof. is proved by using Theorem 6.
Now we are going to apply the same method for solving , where is TFN.
Theorem 12. If is a matrix with crisp coefficients and is a TFN vector the same as , then the system has TFN solution(s), , the same as if and only if the systems (20) have solution: where
Proof. From part (i) of Theorem 5 is a TFN solution of system if and only if So according to part (ii) of Theorem 5 for we have Also according to part (iii) of Theorem 5 for we have but , .
Theorem 13. The system with condition has a solution if and only if optimal solution of the following linear programing is zero:
Proof. This is proved by using Theorem 6.
Here we describe the proposed method completely and step by step by two examples. In the first example, system is introduced in which matrix is a definite (crisp) matrix and is a vector of triangular fuzzy numbers and a solution is then calculated for it. In the second example, an interval system without solution is outlined.
Example 1. Consider the following 4 × 6 fuzzy system in which is a triangular fuzzy vector: To solve this system, we proceed in two successive stages according to Theorem 12.
Stage 1. Find , where is the interval . Therefore, the following system must be solved:
Stage 2. After calculating the intervals in the first stage, search for that satisfies . Therefore, the following system must be solved: Here, the system of the first stage (i.e., system (27)) is solved. It is an interval system, so it must be solved in two substages according to Theorem 10. The first substage is finding the interval length, that is, . Thus, the following system is solved: Here right hand side is and . But according to Theorem 11, the presence of solution (29) is equivalent to the following LP problem: We solve it with the Simplex method [11]. Since the optimal value () is zero, system (29) has the following solution: The second substage is to find the center of . So, the solution of the following system is calculated by common methods in linear algebra: Here right hand side is and . Therefore, the solution of the center of is as follows: Finally, a solution for is as follows according to Theorem 10 and solutions (32) and (33): Now we come back to the second stage. According to Theorem 13, solving system (28) is equivalent to the presence of a solution for the following system: Again, we solve this problem with the Simplex method. Since the optimal value is zero, system (28) has the following solution: According to Theorem 12 and solutions (34) and (38) a final solution for system (26) is as follows: In this example, since different solutions can be obtained in the first or second stage (e.g., in the second substage), other solutions can also be achieved. This is not possible in numerical methods [2–5] or the embedding method [1].
Example 2. Consider the 4 × 6 interval system: Here system (39) is solved. It is an interval system, so it is solved in two stages according to Theorem 10. The first stage is to find the interval length, that is, . Thus, the following system is solved: such that However, according to Theorem 11, the presence of solution (39) is equivalent to the following LP problem: We solve it with the Simplex method. Since the optimal value is not zero, in fact , system (39) has no solution. As a result, system (38) has no solution.
In this paper, we presented a method which is novel for transformed fuzzy system of linear equations. The proposed method is applicable rather than other existing methods. Because the base of this method is linear programming, it can explicitly express the presence or the absence of a solution. In addition, if a solution exists, it expresses the possibility of other solutions. This is not possible in numerical methods based on the embedding method. With a slight change, this method can be used for systems whose right hand side is trapezoidal fuzzy numbers; in previous methods, they did not have this ability. Finding the optimal solutions of fuzzy LP problems is one important application of solving fuzzy linear systems. With this new method, these problems can easily be solved. The presented method can introduce fundamental change in operation research with interval data. It also amends some application of linear programming with fuzzy data as some method in [12–15]. In the numerical solution of fuzzy differential equations, this method provides an explicit method so it is extremely efficient. For example, in three-diagonal matrices, parallel processing techniques can be applied on this method if the number of equations is too high. This method can analytically present the algebraic structure of fuzzy polyhedrons in the same way the representation theorem provides the algebraic structure for crisp polyhedrons. The previous methods are lacking this capability.
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LU
decomposition method for solving fuzzy system of linear equations,” Applied Mathematics and Computation, vol. 172, no. 1, pp. 633–643, 2006. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
M. Dehghan and B. Hashemi, “Iterative solution of fuzzy linear systems,” Applied Mathematics and Computation, vol. 175, no. 1, pp. 645–674, 2006. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
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Copyright © 2014 Reza Kargar et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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The basic characteristics of quadratic functions | StudyPug
Three properties that are universal to all quadratic functions: 1) The graph of a quadratic function is always a parabola that either opens upward or downward (end behavior); 2) The domain of a quadratic function is all real numbers; and 3) The vertex is the lowest point when the parabola opens upwards; while the vertex is the highest point when the parabola opens downward.
Basic Concepts: Factoring trinomials, Solving quadratic equations using the quadratic formula, Completing the square, Shortcut: Vertex formula
Related Concepts: Even and odd functions, What is a polynomial function?, Characteristics of polynomial graphs
Determining the Characteristics of a Quadratic Function Using Various Methods
Determine the following characteristics of the quadratic function
y = -2x^2 + 4x + 6
• Opening of the graph
y-
x-
intercept(s)
• Minimum/Maximum value
Using completing the square
From the graph of the parabola, determine the:
Identifying Characteristics of Quadratic function in General Form:
y = ax^2 + bx+c
y = 2{x^2} - 12x + 10
ii) Sketch the graph.
Identifying Characteristics of Quadratic Functions in Vertex Form:
y = a(x-p)^2 + q
y = 2{\left( {x - 3} \right)^2} - 8
y = ax^2 + bx+c
a(x-p)^2 + q
y = ax^2 + bx+c
a(x-p)^2 + q
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E-mail: Y. Dutta
Tel.:+49-(0)228-73-62229
Currently, I am a Postdoc/associate member at the Hausdorff Center for Mathematics, Bonn. Starting in September 2022, I will be a postdoc at the University of Bonn. My postdoctoral mentor is Prof. Daniel Huybrechts. I graduated from Northwestern University with a PhD thesis written under the supervision of Prof. Mihnea Popa.
Birational geometry, Hyperkähler geometry, generic vanishing theory, Hodge modules and Hodge theory. Here is my CV.
Maximal variation of curves on K3 surfaces. with Daniel Huybrechts. To appear in Tunisian journal of Mathematics.
Generic Vanishing, 1-forms, and Topology of Albanese Maps. with Feng Hao and Yongqiang Liu. Preprint available at arXiv:2104.07074 [math.AG]. Submitted.
Vanishing for Hodge ideals on toric varieties. Math. Nachr. 293 (2020), no. 1 (January). DOI
Effective generation and twisted weak positivity of direct images. with Takumi Murayama (Dec 2017). Algebra and Number Theory, vol. 13, no. 2, 2019, DOI.
On the effective freeness of the direct images of pluricanonical bundles.(Jan 2017) Annales de l'Institut Fourier, vol. 70, no. 4, 2020, DOI.
Some Effective Problems in Algebraic Geometry and their Applications, PhD Thesis
Currently I am not teaching
Wintersemester 2021-22: V5A4 Selected Topics in Algebraic Geometry: Moduli of vector bundles.
Wintersemester 2020-21: S4A1 Graduate Seminar in algebraic geometry: Jacobian of curves and abelian varieties.
Past teaching assistant at Northwestern: Calculus I, II, III, Graph theory, Algebraic Topology I, Introduction to proofs etc.
Notes from my talk at the Bonn/Paris learning serminar on moduli of Hyperkaehler manifolds.
Notes from my talk on D-modules in birational geometry - Short course, Univ. of Chicago, 2018
Notes from my talk in the HPD learning seminar at Bonn. Semi orthogonal decomposition of
\mathrm{D^b\left(}{\mathcal{H}}_{L}\right)
. Supplemental notes computing the cohomology of
{\mathcal{H}}_{L}
GROW@Bonn, a conference aimed at women, non-binary and gender diverse undergraduate students who may be interested in learning about opportunities of graduate degree in the mathematical sciences.
Informal Preprint Seminar at University of Bonn
Hodge Theory and Singularities learning seminar at KU Leuven.
Arbeitsgruppe Komplexe Geometrie, University of Bonn.
Calendar of math events in Bonn.
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A company needs financial capital to operate its business. For most companies, financial capital is raised by issuing debt securities and by selling common stock. The amount of debt and equity that makes up a company’s capital structure has many risk and return implications. Therefore, corporate management must use a thorough and prudent process for establishing a company’s target capital structure. The capital structure is how a firm finances its operations and growth by using different sources of funds.
Empirical Use of Financial Leverage
Financial leverage is the extent to which fixed-income securities and preferred stock are used in a company’s capital structure. Financial leverage has value due to the interest tax shield that is afforded by the U.S. corporate income tax law. The use of financial leverage also has value when the assets that are purchased with the debt capital earn more than the cost of the debt that was used to finance them.
Under both of these circumstances, the use of financial leverage increases the company’s profits. With that said, if the company does not have sufficient taxable income to shield, or if its operating profits are below a critical value, financial leverage will reduce equity value and thus reduce the value of the company.
Given the importance of a company’s capital structure, the first step in the capital decision-making process is for the management of a company to decide how much external capital it will need to raise to operate its business. Once this amount is determined, management needs to examine the financial markets to determine the terms in which the company can raise capital. This step is crucial to the process because the market environment may curtail the ability of the company to issue debt securities or common stock at an attractive level or cost.
With that said, once these questions have been answered, the management of a company can design the appropriate capital structure policy and construct a package of financial instruments that need to be sold to investors. By following this systematic process, management’s financing decision should be implemented according to its long-run strategic plan, and how it wants to grow the company over time.
The use of financial leverage varies greatly by industry and by the business sector. There are many industry sectors in which companies operate with a high degree of financial leverage. Retail stores, airlines, grocery stores, utility companies, and banking institutions are classic examples. Unfortunately, the excessive use of financial leverage by many companies in these sectors has played a paramount role in forcing a lot of them to file for Chapter 11 bankruptcy.
Examples include R.H. Macy (1992), Trans World Airlines (2001), Great Atlantic & Pacific Tea Co. (A&P) (2010), and Midwest Generation (2012). Moreover, excessive use of financial leverage was the primary culprit that led to the U.S. financial crisis between 2007 and 2009. The demise of Lehman Brothers (2008) and a host of other highly levered financial institutions are prime examples of the negative ramifications that are associated with the use of highly levered capital structures.
Financial Leverage In Corporate Capital Structure
Overview of the Modigliani and Miller Theorem on Corporate Capital Structure
The study of a company’s optimal capital structure dates back to 1958 when Franco Modigliani and Merton Miller published their Nobel Prize-winning work “The Cost of Capital, Corporation Finance, and the Theory of Investment.” As an important premise of their work, Modigliani and Miller illustrated that under conditions where corporate income taxes and distress costs are not present in the business environment, the use of financial leverage does not affect the value of the company. This view, known as the Irrelevance Proposition theorem, is one of the most important pieces of academic theory ever published.
Unfortunately, the Irrelevance Theorem, like most Nobel Prize-winning works in economics, requires some impractical assumptions that need to be accepted to apply the theory in a real-world environment. In recognition of this problem, Modigliani and Miller expanded their Irrelevance Proposition theorem to include the impact of corporate income taxes, and the potential impact of distress cost, for purposes of determining the optimal capital structure for a company.
Their revised work, universally known as the Trade-off Theory of capital structure, makes the case that a company’s optimal capital structure should be the prudent balance between the tax benefits that are associated with the use of debt capital, and the costs associated with the potential for bankruptcy for the company. Today, the premise of the Trade-off Theory is the foundation that corporate management should use to determine the optimal capital structure for a company.
Impact of Financial Leverage on Performance
Perhaps the best way to illustrate the positive impact of financial leverage on a company’s financial performance is by providing a simple example. The Return on Equity (ROE) is a popular fundamental used in measuring the profitability of a business as it compares the profit that a company generates in a fiscal year with the money shareholders have invested. After all, the goal of every business is to maximize shareholder wealth, and the ROE is the metric of return on shareholder's investment.
In the table below, an income statement for Company ABC has been generated assuming a capital structure that consists of 100% equity capital. The capital raised was $50 million. Since only equity was issued to raise this amount, the total value of equity is also $50 million. Under this type of structure, the company’s ROE is projected to fall between the range of 15.6% and 23.4%, depending on the level of the company’s pre-tax earnings.
In comparison, when Company ABC’s capital structure is re-engineered to consist of 50% debt capital and 50% equity capital, the company’s ROE increases dramatically to a range that falls between 27.3% and 42.9%.
As you can see from the table below, financial leverage can be used to make the performance of a company look dramatically better than what can be achieved by solely relying on the use of equity capital financing.
Since the management of most companies relies heavily on ROE to measure performance, it is vital to understand the components of ROE to better understand what the metric conveys.
A popular methodology for calculating ROE is the utilization of the DuPont Model. In its most simplistic form, the DuPont Model establishes a quantitative relationship between net income and equity, where a higher multiple reflects stronger performance. However, the DuPont Model also expands upon the general ROE calculation to include three of its parts. These parts include the company’s profit margin, asset turnover, and equity multiplier. Accordingly, this expanded DuPont formula for ROE is as follows:
\begin{aligned} \text{Return on equity} &= \frac{\text{Net Income}}{\text{Equity}}\\ &=\frac{\text{Net Income}}{\text{Sales}} \times \frac{\text{Sales}}{\text{Assets}} \times \frac{\text{Assets}}{\text{Equity}}\\ \end{aligned}
Return on equity=EquityNet Income=SalesNet Income×AssetsSales×EquityAssets
Based on this equation, the DuPont Model illustrates that a company’s ROE can only be improved by increasing the company’s profitability, by increasing its operating efficiency or by increasing its financial leverage.
Measurement of Financial Leverage Risk
Corporate management tends to measure financial leverage by using short-term liquidity ratios and long-term capitalization, or solvency ratios. As the name implies, these ratios are used to measure the ability of the company to meet its short-term obligations. Two of the most utilized short-term liquidity ratios are the current ratio and acid-test ratio. Both of these ratios compare the company’s current assets to its current liabilities.
However, while the current ratio provides an aggregated risk metric, the acid-test ratio provides a better assessment of the composition of the company’s current assets for purposes of meeting its current liability obligations since it excludes inventory from current assets.
Capitalization ratios are also used to measure financial leverage. While many capitalization ratios are used in the industry, two of the most popular metrics are the long-term-debt-to-capitalization ratio and the total-debt-to-capitalization ratio. The use of these ratios is also very important for measuring financial leverage. However, it's easy to distort these ratios if management leases the company’s assets without capitalizing on the assets' value on the company’s balance sheet. Moreover, in a market environment where short-term lending rates are low, management may elect to use short-term debt to fund both its short- and long-term capital needs. Therefore, short-term capitalization metrics also need to be used to conduct a thorough risk analysis.
Coverage ratios are also used to measure financial leverage. The interest coverage ratio, also known as the times-interest-earned ratio, is perhaps the most well-known risk metric. The interest coverage ratio is very important because it indicates a company’s ability to have enough pre-tax operating income to cover the cost of its financial burden.
The funds-from-operations-to-total-debt ratio and the free-operating-cash-flow-to-total-debt ratio are also important risk metrics that are used by corporate management.
Factors Considered in the Capital Structure Decision-Making Process
Many quantitative and qualitative factors need to be taken into account when establishing a company’s capital structure. First, from the standpoint of sales, a company that exhibits high and relatively stable sales activity is in a better position to utilize financial leverage, as compared to a company that has lower and more volatile sales.
Second, in terms of business risk, a company with less operating leverage tends to be able to take on more financial leverage than a company with a high degree of operating leverage.
Third, in terms of growth, faster-growing companies are likely to rely more heavily on the use of financial leverage because these types of companies tend to need more capital at their disposal than their slow growth counterparts.
Fourth, from the standpoint of taxes, a company that is in a higher tax bracket tends to utilize more debt to take advantage of the interest tax shield benefits.
Fifth, a less profitable company tends to use more financial leverage, because a less profitable company is typically not in a strong enough position to finance its business operations from internally generated funds.
The capital structure decision can also be addressed by looking at a host of internal and external factors. First, from the standpoint of management, companies that are run by aggressive leaders tend to use more financial leverage. In this respect, their purpose for using financial leverage is not only to increase the performance of the company but also to help ensure their control of the company.
Second, when times are good, capital can be raised by issuing either stocks or bonds. However, when times are bad, suppliers of capital typically prefer a secured position, which, in turn, puts more emphasis on the use of debt capital. With this in mind, management tends to structure the capital makeup of the company in a manner that will provide flexibility in raising future capital in an ever-changing market environment.
In essence, corporate management utilizes financial leverage primarily to increase the company’s earnings per share and to increase its return-on-equity. However, with these advantages come increased earnings variability and the potential for an increase in the cost of financial distress, perhaps even bankruptcy.
With this in mind, the management of a company should take into account the business risk of the company, the company’s tax position, the financial flexibility of the company’s capital structure, and the company’s degree of managerial aggressiveness when determining the optimal capital structure.
Internal Revenue Service. "Publication 542 (01/2019), Corporations." Accessed May 6, 2020.
Office of Financial Research. "OFR FRAC Working Group: Leveraged Lending & CLOs," Page 15. Accessed May 5, 2020.
Wall Street Journal. "TWA Files for Chapter 11, Approves a Takeover by American Airlines." Accessed May 5, 2020.
UPI. "Macy's Files for Chapter 11." Accessed May 5, 2020.
Proceedings of the National Academy of Sciences of the United States of America. "Measuring the Probability of a Financial Crisis." Accessed May 5, 2020.
Government Publishing Office. "Financial Crisis Inquiry Commission Report," Page 18. Accessed May 5, 2020.
The Nobel Prize. "The Sveriges Riksbank Prize in Economic Sciences in Memory of Alfred Nobel 1985." Accessed May 6, 2020.
The American Economic Review. "The Cost of Capital, Corporation Finance and the Theory of Investment," Page 296. Accessed May 6, 2020.
Murray Frank and Vidhan Goyal. "Handbook of Empirical Corporate Finance: Empirical Corporate Finance: Chapter 12: Trade-Off and Pecking Order Theories of Debt," Pages 140-145. Elsevier. 2011. Accessed May 6, 2020.
New York University. "The Fundamental Determinants of Growth." Accessed May 6, 2020.
Kansas State University Arthur Capper Cooperative Center. "Cooperative Earns, Turns, and Leverage: The DuPont Profitability Model," Page 2. Accessed May 6, 2020.
Webster University. "Financial Ratios." Accessed May 6, 2020.
New York University Stern School of Business. "Financial Ratios and Measures." Accessed May 6, 2020.
New York University Stern School of Business. "Ratings, Interest Coverage Ratios and Default Spread." Accessed May 6, 2020.
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1/7 th of the passenger of a train were children,3/5 th of the passenger were gents and the rest were - Maths - Fractions and Decimals - 7513125 | Meritnation.com
1/7 th of the passenger of a train were children,3/5 th of the passenger were gents and the rest were ladies.If the number of ladies in the train was 360. Find the total number of passengers in the train.
Suppose the total number of passengers in the train is x.
So, number of children =
\frac{1}{7}
th of total passenger =
\frac{x}{7}
Number of gents =
\frac{3}{5}
of total passenger =
\frac{3x}{5}
Then number of ladies = Total passenger - (number of children + number of gents)
⇒360 = x-\left(\frac{x}{7}+\frac{3x}{5}\right)\phantom{\rule{0ex}{0ex}}⇒360 = x-\left(\frac{5x+21x}{35}\right)\phantom{\rule{0ex}{0ex}}⇒360 = x-\frac{26x}{35}\phantom{\rule{0ex}{0ex}}⇒360 = \frac{35x-26x}{35}\phantom{\rule{0ex}{0ex}}⇒\frac{9x}{35} = 360\phantom{\rule{0ex}{0ex}}⇒x = \frac{360×35}{9} = 1400
Therefore total number of passengers in the train is 1400.
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The graph at right shows the profits for a computer company. The horizontal axis represents quarters during a two year period. The vertical axis shows profits in millions of dollars.
During this time period the company had a major failure in its primary computer line. When did this happen?
In the second year, or the fifth quarter.
What appears to be the current trend for the company?
The company is currently experiencing an upswing.
Estimate the net profits for the two year period (start from the first quarter).
13
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Rational connectedness modulo the non-nef locus | EMS Press
Rational connectedness modulo the non-nef locus
It is well known that a smooth projective Fano variety is rationally connected. Recently Zhang [Z2] (and later Hacon and McKernan [HM] as a special case of their work on the Shokurov RC-conjecture) proved that the same conclusion holds for a klt pair
(X,\Delta)
-(K_X+\Delta)
is big and nef. We prove here a natural generalization of the above result by dropping the nefness assumption. Namely we show that a klt pair
(X,\Delta)
-(K_X+\Delta)
is big is rationally connected modulo the non-nef locus of
-(K_X+\Delta)
. This result is a consequence of a more general structure theorem for arbitrary pairs
(X,\Delta)
-(K_X+\Delta)
pseff.
Amaël Broustet, Gianluca Pacienza, Rational connectedness modulo the non-nef locus. Comment. Math. Helv. 86 (2011), no. 3, pp. 593–607
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Battery | Building DC Energy Systems
Building DC Energy Systems
This chapter will cover the necessary basics of electrical batteries in order to understand their usage in a DC energy system. For more detailed information the excellent Battery University website (opens new window) is highly recommended.
# Working Principle
The battery stores chemical energy and can convert it to electrical energy through a reaction. It consists of a cathode (+) and an anode (-) and an electrolyte medium in between.
Figure 1. Basic principle of a battery.
When a consumer such as a lamp or a resistor is connected to the battery, electrons flow from the negative terminal through the consumer to the positive terminal of the battery. This is the opposite direction of the technical current flow direction, which goes from positive to negative portential.
The electrolyte inside the battery allows exchange of ions necessary for the chemical reaction in the electrodes, but it blocks direct electron flow from the anode to the cathode.
# Types of Batteries
Batteries can be classifed into two main categories: non-rechargeable (primary) and rechargeable (secondary). Primary batteries are normally intended for single-use and can not be recharged once depleted. However, they generally store more energy and last longer than rechargeable batteries of same size. The most common primary batteries are Zinc-carbon, Alkaline and Lithium. Secondary batteries can be recharged many times. The most relevant types for stationary energy storage are Lead-acid and Lithium-ion, which will be explained in detail below.
# Lead-acid
Because of their relatively simple design and worldwide availability, lead-acid batteries still have a significant market share for stationary energy storage despite their disadvantages compared to Lithium-ion batteries.
The different types of lead-acid batteries are explained below.
# Flooded lead-acid battery
The flooded or wet battery is the oldest type of lead-acid batteries and not very common anymore. It contains a liquid sulfuric acid electrolyte which moves freely around inside the battery encasement. A reaction occurs between the battery acid and lead plates during the charging and discharging process.
They need to be installed upright in order to make them work properly and avoid leakage of electrolyte. For optimum lifespan, the user must frequently access the cells, check the acid levels and add distilled water as the battery dries out.
# Valve-regulated lead-acid (VRLA) batteries
In contrast to the flooded type, VRLA batteries don't have a liquid electrolyte. It is instead either a gel or absorbed in a glass mat (AGM). The battery enclosure is sealed and contains a valve that is closed under normal conditions. This allows the recombination of the hydrogen and oxygen gases produced and prevents the loss of electrolyte, so the batteries are maintainance-free. If the battery is fast-charging or overcharged and pressure builds up inside the battery, the valve opens and allows some gas to escape.
Gel batteries use a silica additive to convert the previously wet electrolyte into a gel and immobilize it. In contrast to wet batteries, gel batteries don't need to be installed upright anymore to prevent any electrolyte spillage. Chemically, these batteries are still quite similar to flooded lead-acid batteries and don't offer much better performance.
Absorbed Glass Mat (AGM) batteries are the newest type of VRLA batteries. The glass mat construction allows the electrolyte to be absorbed in a thin fiberglass mat sandwiched between the lead plates. This type of battery is more proof against vibrations (important for car batteries) enhancing both the discharge and recharge efficiencies. AGM batteries allow higher currents and offer longer lifespan compared to the other lead-acid batteries.
# Lithium-Ion
The Lithium-ion battery has become the superior battery system on the market in recent years, mainly because of the following advantages:
High energy density, which is important especially for mobile devices like phones and electric vehicles
High number of charging cycles, i.e. long lifespan
Low self-discharge rate and no memory effect, so it does not require deliberate full discharge cycles to maintain its optimum performance.
A major disadvantage of Li-ion batteries is the sensitivity to over-charging or over-discharging, which can result in catastrophic failures including thermal runaway. Hence, it requires a system that monitors its temperature, current and voltage states, protects the battery from operating outside the safe limits in addition to regulating its charging and discharging processes. This so-called Battery Management System (BMS) will be described in a dedicated chapter.
The following table gives a rough overview of different Li-ion technologies and their advantages and disadvantages.
LiNi_xCo_yMn_zO_2
LiNiCoAlO_2
LiFePO_4
NMC or LFP
Anode Graphite Graphite Graphite
Li_4Ti_5O_{12}
Energy ⊕ ⊕ ⊙ ⊖
Power ⊙ ⊖ ⊕ ⊕
Lifetime ⊙ ⊖ ⊕ ⊕
Safety ⊙ ⊖ ⊕ ⊕
For stational storage, LFP (LiFePO4) cells are very suitable due to their comparably good tolerance against overcharge and high cycle life. However, several manufacturers also use NMC cells for stationary storage, that are otherwise mainly used in electric vehicles because of their high energy density.
# Voltage Levels
The voltage of a single electrochemical cell is normally too low to be used in higher power application. If several single cells are connected in series to increase the voltage, this is called battery.
Typical 12V lead-acid batteries consist of 6 cells in series. A battery at a similar voltage using Lithium Iron-Phosphate cells needs only 4 cells, as the single cell voltage is at around 3.3V.
The following interactive graph shows the open circuit voltage (the voltage at the battery terminals without any current flow) for three different types of battery cells vs. their state of charge (SOC).
Li-ion NMC:
Figure 2. Battery voltage vs. state of charge (interactive).
For lithium-ion NMC cells, the open circuit voltage is a good indicator to determine the state of charge of the battery. Lead-acid batteries have a large hysteresis in the open circuit voltage, so the actual voltage measured at the terminal highly depends on whether the battery was charged or discharged before. So the SOC can only be roughly estimated. Lithium iron-phosphate cells have a very flat curve, so the voltage is almost the same at high and low SOC. Thus, additional measures for proper SOC calculation need to be implemented in a battery management system.
# Charge Methods
A good selection of the charging technique will prolong the service lifetime of the batteries, optimize their performance and prevent fatal damages.
# Four-stage battery charging
The charging method with four different stages is only used for lead-acid batteries. Li-ion battery charging is more simple and use only two of the four stages. Figure 3 shows the voltage and current for an entire charging cycle with three or four stages, which will be explained below.
Figure 3. Four battery charging stages.
# 1. Bulk stage
This first charging stage adds about 70 percent of the stored energy. The battery is charged with a constant current (CC), typically the maximum current the charger can supply. In case of a solar charge controller, this is where the Maximum Power Point Tracking (MPPT) takes place. As a result, the terminal voltage increases until the peak charge voltage limit is reached and topping phase is entered.
# 2. Topping stage
During the topping (sometimes also called boost or absorption) the remaining 30 percent of the energy is continued being charged at a constant voltage (CV) while gradually decreasing the charge current until the battery is fully charged.
# 3. Equalization stage
An additional equalization charging is beneficial for flooded lead-acid batteries only. It can be considered as a periodic controlled overcharge that brings the cells to the same charge level and removes the sulphation by increasing the voltage to a higher value than the peak charge voltage limit. The current during equalization is controlled at a very low value.
# 4. Trickle stage
The trickle or float charging stage is the final phase upon the completion of the absorption or equalization phase which maintains the battery at full charge. During this stage, the charge voltage is reduced and held constant. The current reduces to a very low value.
# Two-stage charging (CC-CV)
While lead-acid batteries are charged in 3- or 4-stage profiles, Lithium-ion batteries only go through the bulk and topping charging stages. In the context of Lithium-ion batteries, these two stages are called constant-current (CC) and constant-voltage phases (CV). After the current in the CV phase reached a cut-off threshold, the charging is switched off completely until the voltage drops below a certain recharge threshold.
# Charging Voltages
The following table gives an overview of the target voltages of the different charger stages for different types of 12V batteries. The voltage in brackets is the corresponding single-cell voltage).
Bulk / Topping
Flooded (6 cells) 14.4 V (2.40 V) 15.0 V (2.50 V) 14.1 V (2.35 V)
Gel (6 cells) 14.4 V (2.40 V) n/a 13.8 V (2.30 V)
AGM (6 cells) 14.4 V (2.40 V) n/a 13.8 V (2.30 V)
LiFePO4 (4 cells) 14.2 V (3.55 V) n/a n/a
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Content provided by Libre Solar under CC-BY-SA 4.0 License | Contact / Impressum
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Minimum degree algorithm - Wikipedia
Matrix manipulation algorthim
In numerical analysis, the minimum degree algorithm is an algorithm used to permute the rows and columns of a symmetric sparse matrix before applying the Cholesky decomposition, to reduce the number of non-zeros in the Cholesky factor. This results in reduced storage requirements and means that the Cholesky factor can be applied with fewer arithmetic operations. (Sometimes it may also pertain to an incomplete Cholesky factor used as a preconditioner—for example, in the preconditioned conjugate gradient algorithm.)
Minimum degree algorithms are often used in the finite element method where the reordering of nodes can be carried out depending only on the topology of the mesh, rather than on the coefficients in the partial differential equation, resulting in efficiency savings when the same mesh is used for a variety of coefficient values.
{\displaystyle \mathbf {A} \mathbf {x} =\mathbf {b} }
where A is an
{\displaystyle n\times n}
real symmetric sparse square matrix. The Cholesky factor L will typically suffer 'fill in', that is have more non-zeros than the upper triangle of A. We seek a permutation matrix P, so that the matrix
{\displaystyle \mathbf {P} ^{T}\mathbf {A} \mathbf {P} }
, which is also symmetric, has the least possible fill in its Cholesky factor. We solve the reordered system
{\displaystyle \left(\mathbf {P} ^{T}\mathbf {A} \mathbf {P} \right)\left(\mathbf {P} ^{T}\mathbf {x} \right)=\mathbf {P} ^{T}\mathbf {b} .}
The problem of finding the best ordering is an NP-complete problem and is thus intractable, so heuristic methods are used instead. The minimum degree algorithm is derived from a method first proposed by Markowitz in 1959 for non-symmetric linear programming problems, which is loosely described as follows. At each step in Gaussian elimination row and column permutations are performed so as to minimize the number of off diagonal non-zeros in the pivot row and column. A symmetric version of Markowitz method was described by Tinney and Walker in 1967 and Rose later derived a graph theoretic version of the algorithm where the factorization is only simulated, and this was named the minimum degree algorithm. The graph referred to is the graph with n vertices, with vertices i and j connected by an edge when
{\displaystyle a_{ij}\neq 0}
, and the degree is the degree of the vertices. A crucial aspect of such algorithms is a tie breaking strategy when there is a choice of renumbering resulting in the same degree.
A version of the minimum degree algorithm was implemented in the MATLAB function symmmd (where MMD stands for multiple minimum degree), but has now been superseded by a symmetric approximate multiple minimum degree function symamd, which is faster. This is confirmed by theoretical analysis, which shows that for graphs on n vertices and m edges, MMD has a tight upper bound of
{\displaystyle O(n^{2}m)}
on its run time, whereas for AMD a tight bound of
{\displaystyle O(nm)}
Markowitz, H. M. (1957). "The elimination form of the inverse and its application to linear programming". Management Science. 3 (3): 255–269. doi:10.1287/mnsc.3.3.255. JSTOR 2627454.
George, Alan; Liu, Joseph (1989). "The evolution of the Minimum Degree Ordering Algorithm". SIAM Review. 31 (1): 1–19. doi:10.1137/1031001. JSTOR 2030845.
Tinney, W. F.; Walker, J. W. (1967). "Direct solution of sparse network equations by optimally ordered triangular factorization". Proc. IEEE. 55 (11): 1801–1809. doi:10.1109/PROC.1967.6011.
Rose, D. J. (1972). "A graph-theoretic study of the numerical solution of sparse positive definite systems of linear equations". In Read, R. C. (ed.). Graph Theory and Computing. New York: Academic Press. pp. 183–217. ISBN 0-12-583850-6.
Heggernes, P.; Eisenstat, S. C.; Kumfert, G.; Pothen, A. (December 2001), The Computational Complexity of the Minimum Degree Algorithm (PDF) (Technical report), Institute for Computer Applications in Science and Engineering
Retrieved from "https://en.wikipedia.org/w/index.php?title=Minimum_degree_algorithm&oldid=1087429896"
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This problem is a checkpoint for writing and solving exponential equations. It will be referred to as Checkpoint 9A.
When rabbits were first brought to Australia, they had no natural enemies. There were about
80{,}000
rabbits in
1866
. Two years later, in
1868
, the population had grown to over
2{,}400{,}000
Why would an exponential equation be a better model for this situation than a linear one? Would a sine function be better or worse? Why?
Write an exponential equation for the number of rabbits
t
years after
1866
How many rabbits do you predict were present in
1871
According to your model, in what year was the first pair of rabbits introduced into Australia? Is this reasonable?
24
rabbits were introduced in
1859
, so the model is not perfect, but is close. Is your exponential model useful for predicting how many rabbits there are now? Explain.
Answers and extra practice for the Checkpoint problems are located in the back of your printed textbook or in the Reference Tab of your eBook. If you have an eBook for CCA2, login and then click the following link: Checkpoint 9A: Writing and Solving Exponential Equations
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Gate valve in an isothermal liquid system - MATLAB - MathWorks Italia
Gate Valve (IL)
Valve orifice diameter
Gate position when fully covering orifice
Gate valve in an isothermal liquid system
The Gate Valve (IL) block models flow control by a gate valve in an isothermal liquid network. The valve comprises a round, sharp-edged orifice and a round gate with the same diameter. The gate opens or closes according to the displacement signal at port S. A positive signal retracts the gate and opens the valve.
Gate Valve Opening Schematic
The area open to flow as the gate retracts is:
{A}_{open}=\frac{\pi {d}_{0}^{2}}{4}-{A}_{shielded},
where d0 is the Valve orifice diameter. The area shielded by the gate in a partially open valve is:
{A}_{shielded}=\frac{{d}_{0}^{2}}{2}{\mathrm{cos}}^{-1}\left(\frac{\Delta l}{{d}_{0}}\right)-\frac{\Delta l}{2}\sqrt{{d}_{0}^{2}-{\left(\Delta l\right)}^{2}},
where Δl is the gate displacement, which is the sum of the signal at port S and the Gate position when fully covering orifice.
If displacement exceeds the Orifice diameter, the valve area Aopen is the sum of the maximum orifice area and the Leakage area:
{A}_{open}=\frac{\pi }{4}{d}_{0}^{2}+{A}_{leak}.
For any combination of the signal at port S and the gate offset that is less than 0, the minimum valve area is the Leakage area.
At the extremes of the gate displacement range, you can maintain numerical robustness in your simulation by adjusting the block Smoothing factor. When the smoothing factor is nonzero, a smoothing function is applied to every calculated displacement, but primarily influences the simulation at the extremes of this range.
The normalized gate displacement is:
\Delta \stackrel{^}{l}=\frac{\Delta l}{\Delta {l}_{\mathrm{max}}},
where Δlmax is the Valve orifice diameter.
The Smoothing factor, s, is applied to the normalized displacement:
\Delta {\stackrel{^}{l}}_{smoothed}=\frac{1}{2}+\frac{1}{2}\sqrt{\Delta {\stackrel{^}{l}}_{}^{2}+{\left(\frac{s}{4}\right)}^{2}}-\frac{1}{2}\sqrt{{\left(\Delta \stackrel{^}{l}-1\right)}^{2}+{\left(\frac{s}{4}\right)}^{2}}.
The smoothed displacement is:
\Delta {l}_{smoothed}=\Delta {\stackrel{^}{l}}_{smoothed}\Delta {l}_{\mathrm{max}}.
{\stackrel{˙}{m}}_{A}+{\stackrel{˙}{m}}_{B}=0.
\stackrel{˙}{m}=\frac{{C}_{d}{A}_{valve}\sqrt{2\overline{\rho }}}{\sqrt{P{R}_{loss}\left(1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\right)}}\frac{\Delta p}{{\left[\Delta {p}^{2}+\Delta {p}_{crit}^{2}\right]}^{1/4}},
Avalve is the valve open area.
\overline{\rho }
The critical pressure difference, Δpcrit, is the pressure differential associated with the Critical Reynolds number, Recrit, which is the flow regime transition point between laminar and turbulent flow:
\Delta {p}_{crit}=\frac{\pi \overline{\rho }}{8{A}_{valve}}{\left(\frac{\nu {\mathrm{Re}}_{crit}}{{C}_{d}}\right)}^{2}.
P{R}_{loss}=\frac{\sqrt{1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\left(1-{C}_{d}^{2}\right)}-{C}_{d}\frac{{A}_{valve}}{{A}_{port}}}{\sqrt{1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\left(1-{C}_{d}^{2}\right)}+{C}_{d}\frac{{A}_{valve}}{{A}_{port}}}.
S — Gate displacement, m
Gate displacement, in m, specified as a physical signal. A positive signal retracts the gate and opens the valve.
Valve orifice diameter — Orifice diameter
Valve open-area diameter.
Gate position when fully covering orifice — Gate offset
Gate offset when the valve is closed. A positive, nonzero value indicates a partially open valve. A negative, nonzero value indicates an overlapped valve that remains closed for an initial displacement set by the physical signal at port S.
Areas at the entry and exit ports A and B, which are used in the pressure-flow rate equation that determines the mass flow rate through the orifice.
Needle Valve (IL) | Poppet Valve (IL) | Orifice (IL)
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Transformer is a passive electrical device used to increase or decrease AC voltage. It is based upon principle of electromagnetic induction. There are two coils on transformer. The coil on which power in fed is called primary coil and that one from which output power is taken is called secondary coil. Both coils are wound on laminated iron to reduce loss of electric energy by heating due to eddy current. Transformer cannot change the voltage of DC current. There are two types of transformers regarding to voltage level
Step Up Transformer: It is used to increase voltage of AC current. In step up transformer number of turns on secondary coil is greater than that of primary coil.
Step down Transformer: It is used to decrease the voltage of AC current. In step down transformer number of turns in primary coil is greater tha that in secondary coil.
The value of the power for an electric circuit is the value of the voltage by the value of the current intensity. As in the case of a transformer, the value of the power in the primary is the same value for the power in the secondary we have:
(input voltage on the primary coil) × (input current on the primary coil ) = (output voltage on the secondary coil ) × (output current on the secondary coil)
This can be written as an equation:
{V}_{p}
{I}_{p}
{V}_{s}
{I}_{s}
We can also work out the transformer output voltage if we know the input voltage and the number of turns on the primary and secondary coils.
\frac{inputvoltageontheprimarycoil}{outputvoltageonthesecondarycoil}
\frac{numberofturnsofwireontheprimarycoil}{numberofturnsofwireonthesecondarycoil}
\frac{{V}_{p}}{{V}_{s}}
\frac{{n}_{p}}{{n}_{s}}
{V}_{p}
input voltage on the primary coil.
{V}_{s}
input voltage on the secondary coil.
{I}_{p}
input current on the primary coil.
{I}_{s}
input current on the secondary coil.
{n}_{p}
the number of turns of wire on the primary coil.
{n}_{s}
the number of turns of wire on the secondary coil.
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Understanding scale drawings with examples | StudyPug
Diagrams that you see may not always be drawn to its intended exact size. Instead, it may be scaled up to help you see the diagram better, or it may be scaled down so that it can fit on the medium that you're seeing the diagram on. Whichever it may be, scale diagrams may be drawn to a certain scale factor.
Scale factors are the ratio of two corresponding measurements on two similar figures. It is used in scaling and isn't constrained to 2D shapes. You can also use a scale factor with 3D shapes. If you have the scale factor, you can convert a measurement into another similar figure's measurements. An example of this in use is when you're changing a diagram on a paper's measurements to the real-life item.
Let's say you've got two figures: figure A and figure B. They are both similar, and figure B is an enlargement of figure A. How do you go about finding the scale factor between the two?
The scale factor, as a reminder, is a ratio, so what we're looking for is a ratio that can relate figure A and B. You can calculate the scale factor when you've got an enlargement by putting a side of the bigger figure divided by the smaller figure. That is:
In this case, it would mean that the scale factor of the enlargement =
\frac{B side}{A side}
Alternatively, if you were looking at the scale factor of a reduction you'd put the smaller figure's side over the bigger figure's side. That is:
See if you can tackle the below scale factor examples that deal with scale drawings and put into use what you know about the scale factor definition!
What is the actual length of this object?
The scale for the image of a leaf is 1 : 0.2.
The scale for the image of a leaf
ratio = \frac{image}{actual} = \frac{1}{0.2} = 50cm
Find the unknown:
\frac{1 \times 50}{0.2 \times 50} = \frac{50cm}{10cm}
10cm is the actual length of the leaf.
The scale for the image of a building is 1: 406
The scale for the image of a building
ratio = \frac{image}{actual} = \frac{1}{406} = 45cm
\frac{1 \times 45}{406 \times 45} = \frac{45cm}{18270cm}
18270cm is the actual height of the building.
Tony is 6cm tall in a photo. Yet, his actual height is 1.86 m. What is the scale factor?
scale factor = \frac{image}{actual}
First, we need to convert unit:
1.86m \times \frac{100cm}{1m} = 186cm
So, put numbers in
scale factor = \frac{image}{actual}
scale factor = \frac{6cm}{186cm}
Simplify it and we get the final answer
\frac{1}{31}
Simplify scale factor
Want to double check your answers? Feel free to use this scale conversion calculator to help you out.
Coming up next, building on what you learn in this chapter, you'll be learning about congruent triangles and proving their congruence via SSS, SAS, HL, ASA, and AAS.
What is the actual length of a street on the map? All we need is a ruler and the map's scale factor! By using the scale factors, we can easily find the actual size of an object. See how to do that in this lesson.
Basic Concepts: Ratios, Enlargements and reductions with scale factors
What is the actual length of each object?
The scale for the image of a leave is 1: 0.2.
Tony is 6 cm tall in a photo. Yet, his actual height is 1.86 m. What is the scale factor?
A spaceship is 56 m in height. You have built a model of the spaceship with a scale factor of 1:60 for your cousin overseas. However, the longest shipping box you can find is only 0.8 m. Will the spaceship model fit into the shipping box?
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Margin Trading and Securities Lending, Investor Sentiments and the Volatility of Chinese Securities Market
\sigma \text{=}\sqrt{\frac{1}{4N\mathrm{ln}2}{\sum }_{t=1}^{N}{\left(\mathrm{ln}\frac{{h}_{t}}{{l}_{t}}\right)}^{2}}
{h}_{t}
{l}_{t}
\sigma =\sqrt{\frac{1}{N}{\sum }_{i=1}^{N}\left[\left(\mathrm{ln}\frac{{h}_{i}}{{l}_{i}}\right)\left(\mathrm{ln}\frac{{h}_{i}}{{o}_{i}}\right)+\left(\mathrm{ln}\frac{{l}_{i}}{{c}_{i}}\right)\left(\mathrm{ln}\frac{{l}_{i}}{{o}_{i}}\right)\right]}
{h}_{t}
{l}_{t}
{c}_{i}
{o}_{i}
{\sigma }^{2}={\alpha }_{0}+{\sum }_{i=1}^{p}{\alpha }_{i}{\sigma }_{t-i}^{2}+{\sum }_{j=1}^{q}{\beta }_{j}{\mu }_{t-j}^{2}
\mathrm{ln}{\sigma }_{t}^{2}={\alpha }_{0}+{\sum }_{i=1}^{p}{\alpha }_{i}\mathrm{ln}{\sigma }_{t-i}^{2}+{\sum }_{j=1}^{p}{\beta }_{j}\left[\theta {z}_{t-j}+\gamma \left(|{z}_{t-j}|-E|{z}_{t-j}|\right)\right]
{z}_{t}=\frac{{\mu }_{t}}{{\sigma }_{t}}
{R}_{t}={\alpha }_{0}+{\alpha }_{1}{R}_{t-1}+{\alpha }_{2}{R}_{t}^{IF00}+{\alpha }_{3}{\sigma }_{t}^{2}+{u}_{t}
{u}_{t}={\sum }_{i=1}^{p}{\rho }_{i}{u}_{t-i}+{\sum }_{j=1}^{q}{\theta }_{j}{\epsilon }_{t-j}+{\epsilon }_{t}
\begin{array}{l}\mathrm{ln}\left({\sigma }_{t}^{2}\right)={\beta }_{0}+{\sum }_{k=1}^{m}{\varphi }_{i}\mathrm{ln}\left({\sigma }_{t-k}^{2}\right)+{\sum }_{l=1}^{n}{\phi }_{l}|{z}_{t-l}|+{\gamma }_{1}{z}_{t-1}+{\gamma }_{2}{N}_{t}+{\gamma }_{3}{D}_{t}^{bull}\\ \text{}+{\gamma }_{4}{P}_{t}+{\gamma }_{5}{V}_{t}+{\gamma }_{6}{Q}_{t}^{mr}+\text{ }{\gamma }_{7}{Q}_{t}^{ss}+{\gamma }_{8}{Q}_{t}^{IF00}+{\gamma }_{9}{D}_{t}^{bull}{P}_{t}\\ \text{}+{\gamma }_{10}{D}_{t}^{bull}{V}_{t}+{\gamma }_{11}{D}_{t}^{bull}{Q}_{t}^{mr}+{\gamma }_{12}{D}_{t}^{bull}{Q}_{t}^{ss}+{\gamma }_{13}{D}_{t}^{bull}{Q}_{t}^{IF00}\end{array}
{R}_{t}
t
{R}_{t-1}
{\sigma }_{t}^{2}
{u}_{t}
{N}_{t}
{R}_{t}^{IF100}
t
{D}^{bull}
{P}_{t}
{V}_{t}
{Q}_{t}^{MR}
t
{Q}_{t}^{SS}
t
{Q}_{t}^{MR}
{Q}_{t}^{SS}
{Q}_{t}^{IF00}
t
{Q}_{t}^{IF00}
{\gamma }_{1},\cdots ,{\gamma }_{13}
{\alpha }_{1}
{\alpha }_{2}
\rho
\theta
\phi
{N}_{t}
Huang, H.T. (2019) Margin Trading and Securities Lending, Investor Sentiments and the Volatility of Chinese Securities Market. American Journal of Industrial and Business Management, 9, 536-550. https://doi.org/10.4236/ajibm.2019.93036
2. Li, K., Xu, L. and Zhu, W. (2014) Short-Sale Constrains and Stock Mispricing: The Evidence from the Margin Transactions Institution. Journal of Economic Research, 10, 166-178.
3. Li, Z., Du, S. and Lin, B. (2015) Short Selling and Stock Price Stability: Natural Experiments from China’s Margin Financing Market. Journal of Financial Research, 6, 173-188.
4. Xiao, H. and Kong, A. (2014) Study on the Influence Mechanism of Margin Financing on Stock Price Traits Fluctuation: Test Based on Double Difference Model. Management World, 8, 30-43.
5. Chen, H. and Fan, Y. (2015) Financing and Margin Trading System vs. China Stock Market Volatility: An Analysis Based on Panel Data Policy Evaluation Method. Journal of Financial Research, 6, 159-172.
6. Xu, H. and Chen, X. (2012) Does China’s Launch of Margin Financing and Securities Lending Transactions Promote the Pricing Efficiency of the Underlying Stocks? Management World, 5, 52-61.
7. Li, F. (2017) Does Margin Trading Aggravate Stock Market Fluctuation? From the Perspective of Asymmetric Volatility. Journal of Financial Research, 2, 147-162.
8. Lin, X., Yi, F. and Chen, C. (2016) The Effect of Marginal and Financing Trading: A Study Based on the Double Difference Model. Investment Research, 4, 74-86.
9. Zhu, J. and Fang, J. (2016) Chinese-style Margin Financing System Arrangement and Deterioration of Stock Price Crash Risk. Economic Research Journal, 5, 143-158.
10. Black, F. (1976) Studies of Stock Price Volatility Changes. Proceedings of the 1976 Meetings of the Business and Economics Statistics Section, American Statistical Association, Washington DC, 177-181.
14. Parkinson, M. (1980) The Extreme Value Method for Estimating the Variance of the Rate of Return. Journal of Business, 53, 61-65. https://doi.org/10.1086/296071
15. Rogers, L.C.G., Satchell, S.E. and Yoon, Y. (1994) Estimating the volatility of stock Prices: A comparison of Methods That Use High and Low Prices. Applied Financial Economics, 4, 241-247. https://doi.org/10.1080/758526905
16. Tong, B., Zhang, C. and Xiao, L. (2013) Volatility Trading. Shanghai Jiaotong University Press, Shanghai.
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Chakra Control (Ninjas) - Ninja Fantasy
We begin with the explanation of the chakra control: The chakra is the vital energy of the body which is essential for any jutsu, through concentration and control of the chakra new types of chakra can be formed. The most common product is the elemental chakra, which is used to perform elemental jutsu. This is why each of our ninja manages a type of elemental chakra (Fire, Wind, Lightning, Earth, Water), during Chakra Control V-1.0 the mission is to concentrate that ninja's energy to obtain a daily reward which will be in NFS tokens. This will basically work as follows:
Each ninja has a rank, which defines the fighting level, and other stast within the game but also defines the control of the chakra so that some ninjas can accumulate more chakra and receive more rewards. As our number of graduated ninjas is limited (1100) these are the ninjas that can accumulate more chakra and therefore more rewards, being the kages the ones that accumulate more rewards daily. But also as we know that 1100 ninjas is not enough for people who want to acquire an NFT, we have designed some low-level ninjas that can also enter the status of chakra control, these are the aspirants, you can only have a maximum of 10 aspirants per account and multi-accounts are not allowed. All ninjas can enter chakra control, but be careful the maximum number of applicants is also limited.
How to enter chakra control?
The first thing is to register in our marketplace, and have an NFT to activate the chakra mode. If you already have it, you can go to the My NFTs page in the market and under each NFT you will see the button to activate the chakra mode along with the fees to start. You must have the NFS defined by the oracle in your market wallet, and a little BNB for fees, within the chakra mode to perform the tasks you will not need BNB.
If you don't have NFT, you can buy an aspirants in our market. For that you must register, deposit the NFS for the scrolls, and a little bnb for fees and that you can minted the NFT to receive royalties. After buying the scroll, you can see it on the My NFTs page, and open the scroll and you will receive your ninja. You will see the amount to pay in NFS to activate the chakra mode and you can activate it in each ninja.
Duration of the fees for the chakra mode?
You must pay a fee in NFS defined by the oracle to activate the chakra mode, this fee will keep your ninja in chakra mode for 7 days and after that you must pay again to activate it.
How do I start receiving rewards?
Once you have your NFT in chakra control status, daily you must complete some tasks, these tasks will be given by a sensei (guide NPC) and will appear daily on your panel for you to complete and receive your rewards. Very important, be careful and complete your tasks because the ninja can lose concentration and not generate chakra during the day, which means that you will not receive rewards.
Every 24 hours the rewards will accumulate in your panel, this will be updated daily at 00:00 UTC. Every 7 days the reward of day 1 is released, and if you accumulate each reward, the tax to withdraw is reduced. It starts at 28% tax on day 7, and is reduced by 4% daily until having 0% of fees on day 15.
You can have a maximum number of 10 ninjas per account, and you can place them all in chakra control status. We want to be clear about this, because we want to avoid multi-accounts, so we give a margin of 10 ninjas per account so that there is no need to open multiple accounts. If anyone abuses this, we will have to ban the accounts.
What profit will I get with the chakra control?
Currently we cannot give an exact number because this depends on the rank of the ninja you have and the amount of tokens is defined by our AI based on: the price of the token, the number of ninjas in a status of chakra control, the remaining available supply of tokens for the chakra control, the estimated time to launch the game, the daily time spent in control chakra, and of course completing the daily tasks.
The rewards are managed per oracle, as are the daily rewards based on the token price in BUSD.
The chakra mode started on November 5, you can now buy your ninja in the marketplace and enter chakra mode to start collecting rewards.
The rewards are counted daily, each user can see the rewards of their ninjas but they can only be withdrawn after 7 days of being in this chakra control status.
The rewards depend on the rank of each ninja, and are defined by our AI, based on the following variables: The price of the token, the number of ninjas in chakra control status, the remaining available supply for the control chakra and the estimated time for the launch of the game.
Table of rewards estimated by one of our partners: LINK
For all players who want to have more ninjas in the chakra control, you can buy additional slots to be able to place the desired amount of ninjas, thus you will not need to create multiple accounts that will be banned by the system.
We are introducing our algorithm as control mechanisms to define the amount of tokens that are delivered daily, taking into account the dollar price of the token as the main factor.
The chakra control reward system is based on a simple interest formula, with some modifications.
F ( c ,t )=c ( 1+p t )
where (c), is the amount of tokens available for chakra control, (p) is the daily interest, (t) is the chakra control period in days and (F) is the future value of (c). Since that in Ninja Fantasy ecosystem we continuously we need to modify the amount of rewards, to give more entry to new users, maintain the estimated time for the launch of the game, and control inflation, we have to reduce the amount of rewards at certain times since the total supply for chakra control could arrive to 0 fast.
The total appreciation principle:
R T (c,t)=F(c,t)−c=cp e t
The chakra control rewards being calculated and locked at the beginning of every reward cycle, which is approximately 30 days, and the total reward itself is set to 20% of total funds in the treasury. That said, the total appreciation principle allocated from that portion
T
and therefore its a subject to:
R T (c,t)=cp e t≤T
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Kemnitz's conjecture - Wikipedia
Kemnitz's conjecture
Every set of lattice points in the plane has a large subset with centroid a lattice point
In additive number theory, Kemnitz's conjecture states that every set of lattice points in the plane has a large subset whose centroid is also a lattice point. It was proved independently in the autumn of 2003 by Christian Reiher, then an undergraduate student, and Carlos di Fiore, then a high school student.[1]
The exact formulation of this conjecture is as follows:
{\displaystyle n}
be a natural number and
{\displaystyle S}
a set of
{\displaystyle 4n-3}
lattice points in plane. Then there exists a subset
{\displaystyle S_{1}\subseteq S}
{\displaystyle n}
points such that the centroid of all points from
{\displaystyle S_{1}}
is also a lattice point.
Kemnitz's conjecture was formulated in 1983 by Arnfried Kemnitz[2] as a generalization of the Erdős–Ginzburg–Ziv theorem, an analogous one-dimensional result stating that every
{\displaystyle 2n-1}
integers have a subset of size
{\displaystyle n}
whose average is an integer.[3] In 2000, Lajos Rónyai proved a weakened form of Kemnitz's conjecture for sets with
{\displaystyle 4n-2}
lattice points.[4] Then, in 2003, Christian Reiher proved the full conjecture using the Chevalley–Warning theorem.[5]
^ Savchev, S.; Chen, F. (2005). "Kemnitz' conjecture revisited". Discrete Mathematics. 297 (1–3): 196–201. doi:10.1016/j.disc.2005.02.018.
^ Kemnitz, A. (1983). "On a lattice point problem". Ars Combinatoria. 16b: 151–160.
^ Erdős, P.; Ginzburg, A.; Ziv, A. (1961). "Theorem in additive number theory". Bull. Research Council Israel. 10F: 41–43.
^ Rónyai, L. (2000). "On a conjecture of Kemnitz". Combinatorica. 20 (4): 569–573. doi:10.1007/s004930070008.
^ Reiher, Ch. (2007). "On Kemnitz' conjecture concerning lattice-points in the plane". The Ramanujan Journal. 13: 333–337. arXiv:1603.06161. doi:10.1007/s11139-006-0256-y.
Gao, W. D.; Thangadurai, R. (2004). "A variant of Kemnitz Conjecture". Journal of Combinatorial Theory. Series A. 107 (1): 69–86. doi:10.1016/j.jcta.2004.03.009.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Kemnitz%27s_conjecture&oldid=1032224178"
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Chang Young was attempting to evaluate the following area:
\int _ { 1 } ^ { 5 } ( 5 x + 2 ) d x
He showed the following steps:
\left. \begin{array} { l } { \frac { 5 } { 2 } 5 ^ { 2 } + 2 ( 5 ) + C - \frac { 5 } { 2 } 1 ^ { 2 } + 2 ( 1 ) + C } \\ { \frac { 5 } { 2 } \cdot 25 + 10 + C - \frac { 5 } { 2 } + 2 + C } \\ { \frac { 125 } { 2 } - \frac { 5 } { 2 } + 12 + 2 C } \\ { \frac { 120 } { 2 } + 12 + 2 C } \\ { 60 + 12 + 2 C } \\ { 72 + 2 C } \end{array} \right.
He knows that this is a definite integral and there should not be any
C^\prime
s. Also, the teacher said the answer was
68
. He needs your help to find his error and find out how to eliminate his
+ 2C
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Charles C. Pugh - Wikipedia
This article is about the mathematician. For other people with the same name, see Charles Pugh (disambiguation).
Charles Chapman Pugh (born 1940) is an American mathematician who researches dynamical systems. Pugh received his PhD under Philip Hartman of Johns Hopkins University in 1965, with the dissertation The Closing Lemma for Dimensions Two and Three.[1] He has since been a professor, now emeritus, at the University of California, Berkeley.
Charles Pugh, Berkeley, 1993
Work in dynamical systems
The Closing Lemma for Dimensions Two and Three (1965)
https://math.berkeley.edu/people/faculty/charles-c-pugh
In 1967 he published a closing lemma named after him in the theory of dynamical systems.[2][3] The lemma states: Let f be a diffeomorphism of a compact manifold with a nonwandering point x.[4] Then there is (in the space of diffeomorphisms, equipped with the
{\displaystyle C^{1}}
topology) in a neighborhood of f a diffeomorphism g for which x is a periodic point. That is, by a small perturbation of the original dynamical system, a system with periodic trajectory can be generated.
In 1970 he was an invited speaker at the International Congress of Mathematicians in Nice, delivering a talk on Invariant Manifolds.
Mary Cartwright (left) with Charles Pugh, Nice, 1970
Real Mathematical Analysis, Springer-Verlag, 2002
^ Charles C. Pugh at the Mathematics Genealogy Project
^ Bonatti, Christian (June 10, 2008). "Pugh closing lemma". Scholarpedia. 3 (6): 5072. doi:10.4249/scholarpedia.5072. ISSN 1941-6016.
^ Pugh, Charles C. (1967). "An Improved Closing Lemma and a General Density Theorem". American Journal of Mathematics. 89 (4): 1010–1021. doi:10.2307/2373414. ISSN 0002-9327. JSTOR 2373414.
^ Wandering points were introduced by George Birkhoff to describe dissipative systems (with chaotic behavior). In the case of a dynamical system given by a map f, a point wanders if it has a neighborhood U which is disjoint to all of the iterations of the map on it:
{\displaystyle f^{n}(U)\cap U=\varnothing .\,}
Retrieved from "https://en.wikipedia.org/w/index.php?title=Charles_C._Pugh&oldid=1054748520"
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Experimental and Numerical Investigations on the Leakage Flow Characteristics of Helical-Labyrinth-Brush Seals | J. Eng. Gas Turbines Power | ASME Digital Collection
Shanghai Electric Gas Turbine Co. Ltd.,
e-mail: heyuan@shanghai-electric.com
Jingjin Ji,
Jingjin Ji
e-mail: jijj@shanghai-electric.com
e-mail: sunbo@shanghai-electric.com
Zhang, Y., Li, J., Ma, D., He, Y., Ji, J., Sun, B., Li, Z., and Yan, X. (March 10, 2021). "Experimental and Numerical Investigations on the Leakage Flow Characteristics of Helical-Labyrinth-Brush Seals." ASME. J. Eng. Gas Turbines Power. April 2021; 143(4): 041023. https://doi.org/10.1115/1.4048842
The helical-labyrinth seal (HLS) can reduce rub between labyrinth teeth and rotor during the rotor vibration because the helical teeth on the stator and steps (or teeth) on the rotor are staggered in some positions. The helical-labyrinth seal with the bristle pack named as the helical-labyrinth-brush seal (HLBS) has excellent sealing performance, but the study on the leakage flow characteristics of the HLBS is not available. This paper, using computational fluid dynamic (CFD) analysis technology based on a porous medium model, investigates the leakage flow characteristics of two types of HLBSs (bristle pack installed upstream or downstream of helical-labyrinth tooth named as HLBS-U and HLBS-D, respectively) at various pressure ratios (1–1.3) and rotational speeds (0–10,000 r/min, surface speeds range from 0 to 209 m/s). The radial clearance
cb
between the rotor and the bristle pack ranges from 0 mm to 1.0 mm, and the radial clearance
ct
between the labyrinth teeth and the steps on the rotor is 1.6 mm. In parallel, the leakage flow characteristics of the HLBS-D with the constant
cb
of 1.0 mm are experimentally measured at the pressure ratio up to 1.3 and rotational speed up to 2000 r/min (surface speed 42 m/s). The CFD-derived leakage flow rate (represented as effective clearance) and static cavity pressure agree well with the experimental data in the whole range of test conditions. The shaft rotation eliminates the leakage hysteresis effect of the HLBS-D. Compared with the HLBS-D, the effective clearance of HLBS-U is less sensitive to rotational speed changes. The effective clearance of the HLBS-U is smaller than that of the HLBS-D in the case of
cb
= 0.5 mm and rotational speed n < 10,000 r/min, and the case of
cb
= 1.0 mm. However, for the case of
cb
= 0.5 mm and n = 10,000 r/min, and the case of
cb
≤ 0.1 mm, the situation is opposite. The brush seal sections of the HLBS-U and the HLBS-D offer over 55% and 65% total static pressure drop in the case of
cb
= 1.0 mm, respectively. The brush seal sections of two HLBSs bear almost the same static pressure drop of the over 97% total static pressure drop as
cb
equals to 0.1 mm. What is more, the HLBS-U has lower turbulent kinetic energy upstream of the bristle pack than the HLBS-D does, which means that the intensity of the bristles flutter of the HLBS-U is lower.
Cavities, Clearances (Engineering), Computational fluid dynamics, Flow (Dynamics), Leakage, Leakage flows, Porous materials, Pressure, Rotors, Pressure drop
Suerken
Effects of Tooth Bending Damage on the Leakage Performance and Rotordynamic Coefficients of Labyrinth Seals
Ozmusul
Leakage and Cavity Pressures in an Interlocking Labyrinth Gas Seal: Measurements Versus Predictions
Axial Inclination of the Bristles Pack, a New Design Parameter of Brush Seals for Improved Operational Behavior in Steam Turbines
Polklas
Generalizing Circular Brush Seal Leakage Through a Randomly Distributed Bristle Bed
Numerical Modeling of Flows in Simulated Brush Seal Configurations
Paper No. AIAA-90-2141.10.2514/6.AIAA-90-2141
Hybrid Brush Pocket Damper Seals for Turbomachinery
Bowshe
Application of Blade Tip Shroud Brush Seal to Improve the Aerodynamic Performance of Turbine Stage
Experimental and Numerical Investigations on Leakage Flow Characteristics of Two Kinds of Brush Seals
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106.3.2.80 TM-80, Measurement of Retroreflectivity by Handheld Retroreflectometers - Engineering_Policy_Guide
This method describes retroreflectivity inspections of pavement markings using a handheld retroreflectometer.
1 106.3.2.80.1 Equipment
3.1 106.3.2.80.3.1 Solid Longitudinal Lines
3.1.1 106.3.2.80.3.1.1 Road Lengths
{\displaystyle \leq \;}
1000 Ft.
3.1.2 106.3.2.80.3.1.2 1000 Feet < Road Lengths
{\displaystyle \leq \;}
One Mile
3.2 106.3.2.80.3.2 Intermittent Longitudinal Lines
{\displaystyle \leq \;}
{\displaystyle \leq \;}
3.3 106.3.2.80.3.3 Legends, Symbols, Pedestrian Crossing, Etc.
4 106.3.2.80.4 Acceptance
106.3.2.80.1 Equipment
The apparatus shall consist of a handheld 30-meter geometry retroreflectometer, such as the Mirolux 30 retroreflectometer.
Take the pavement marking measurements between seven and fourteen days after application. Waiting seven days ensures that all excess glass spheres have been removed, and completing the measurements by fourteen days allow retest of the pavement marking within 45 days if the handheld reflectometer measurements are challenged.
Take all measurements in the sampling areas in the direction of traffic flow, except on the centerline of two-lane roads, then take the required number of measurements in each direction. Measure both single and double lines, and apply the acceptance criteria for each line in both directions.
Because handheld retroreflectometers normally shoot a beam of light approximately 6-12 in. ahead of the machine’s lens, locate the lens approximately 6-12 in. from the flat area of the rumble strip. When the light falls in the bottom of a rumble strip, the retroreflectivity readings are usually lower than the backside or top of the rumble area because of the change of light beam angle, and for that reason, take care to avoid taking readings in the bottom of the rumble strip.
106.3.2.80.3.1 Solid Longitudinal Lines
106.3.2.80.3.1.1 Road Lengths
{\displaystyle \leq \;}
Randomly select one 300 ft. section of pavement marking for testing. Take measurements approximately every 15 ft. along the pavement marking line. Take a total of 20 measurements.
106.3.2.80.3.1.2 1000 Feet < Road Lengths
{\displaystyle \leq \;}
106.3.2.80.3.2 Intermittent Longitudinal Lines
{\displaystyle \leq \;}
{\displaystyle \leq \;}
106.3.2.80.3.3 Legends, Symbols, Pedestrian Crossing, Etc.
Retroreflectivity acceptance requirements do not apply to intersection markings, such as stop lines, arrows, legends, pedestrian crossings, symbols, etc. No readings are required. Acceptance is based upon quality of work in accordance with Sec 620.50.3.6.
106.3.2.80.4 Acceptance
Use the number of readings below the minimum accepted values as set in Sec 620 to determine if a pavement marking is accepted or rejected.
For cases in which six or three sample size measurements are taken, calculate the average. If the average falls below the minimum required value, reject the pavement marking.
For cases in which 20 sample measurements are taken for a segment of roadway, if 17 or more readings in the 300 ft. section, or actual length for intermittent longitudinal lines less than 1000 ft., for that segment of roadway are above the minimum value required, accept that segment of pavement marking. If four or more readings within a segment are below the minimum value required, calculate the average value of the failed readings and use the calculated average to determine, if applicable, the amount of deduct that applies to that segment, or if that segment of pavement marking needs to be removed and replaced.
Retrieved from "https://epg.modot.org/index.php?title=106.3.2.80_TM-80,_Measurement_of_Retroreflectivity_by_Handheld_Retroreflectometers&oldid=37516"
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Stem-and-leaf plots | StudyPug
To continue with our topics on data representation, it is time to introduce a new approach for graphing data which was created a few centuries ago to accommodate for the use of typewriters. Before computers and high definition graphics existed and were in the mainstream media, statisticians, and researchers in general, wanting to publish data graphic representations to communicate with broader audiences, came up with this clever method called the stem and leaf plot, which allowed graphics to be produced with regular typewriters.
Such method opened up the communication ways for data analysis by not requiring graphic art to represent data, just a simple table-style representation which not only was easily built while typewriting and emulated a histograms representation, but it could actually keep more details from the data being plotted than the frequency distribution representations we have seen before. Thus, out lesson of today focuses on stem and leaf plot statistics.
What is a stem and leaf plot
The stem and leaf plot is another way to represent data in a table-like manner. Stem and leaf plots may be not as intuitive as a frequency distribution table, but they do have advantages against them (which we will cover a little later), for the moment, let us look at the formal stem and leaf plot definition:
A stem and leaf plot is an organized, and position-conscious way to represent data that allows the observer to interpret all the characteristics of the data efficiently. On this kind of plot, the data is split in a two column table; the left column is the stem which contains the leftmost digits or highest multiples of ten in the data set values; then, the right column of the table is the stem and it provides a list of the rightmost digits of each value from the data set.
The concept of a stem and leaf plot of data may sound confusing, therefore, is better if you take a look at the next example: Having the data set: 7, 10, 26, 34, 19, 31, 27, 4, 12, 15, a simple stem and leaf plot would look like:
Figure 1: Stem and leaf plot
We have 4 different tens in the data set: 10 × 0 = 0, 10 × 1 = 10, 10 × 2 = 20, 10 × 3=30. Now notice the column on the left, or stem, it contains the digits that represent each ten from the data set: 0, 1, 2, 3. On the other side, the right column provides a space for each of the remaining digits for each ten of the data set; in other words, if you have the fourth ten in this case, which is 30, the remaining digits are 1 and 4, because in the data set you have a 31 and a 34. Notice the pattern in the plot, the left column represents the tens, and each digit on the right represents the number to be added to each particular ten to form the values from the data set.
This is a basic stem and leaf plot: tens on the left, single digits on the right, but it is not the only way how one can do stem and leaf plots, you can make these same kind of plots with hundreds, thousands, or even decimals! The options are endless, you are the one to decide which tens (or multiples of ten) go on the stem and then just add the appropriate remaining digits on the leaf.
Look at the next examples:
Figure 2: Types of Stem and Leaf plots
Let us explain each of these examples in figure 2 starting with the plot containing the thousands. Notice on the stem column, you have the tens which in this case they refer to digits that are multiples of 10 in the scale of thousands; therefore, each digit on the stem represents the number multiplied by a thousand: 1 × 1000 = 1000, 2 × 1000 = 2000. Then on the leaf column each set of three numbers are the remaining digits for each value in the data set on one of the two thousands ranges.
This particular plot could be done in several other ways. Remember we said that you could pick the tens (or multiples of ten) you select to be part of your stem, thus, the plot in the scale of thousands has many other options you could pick as the multiples of ten in the stem part; this is one we saw right away:
Figure 3: Other ways to write down a stem leaf plot
Notice that in the second notation form of the thousands plot, the stems are each of 3 digits. They still represent multiples of 10: 199 × 10 = 1990, 200 × 10 = 2000, 201 × 10 = 2010; therefore, you just need to add the values of the leaf column (or the leaves) to the stems and you will obtain each of the values of the data set. Both notations of this diagram are valid, and so, when constructing a stem and leaf plot with thousands (or with any values) you can choose the way is most convenient to you.
Going back to figure 2, taking a look at the stem and leaf plot with hundreds: each of the values on the stem column represent itself multiplied by a hundred, which added to the leaves produce values of the data set; hence, in this stem and leaf plot the stems are multiples of 10, not tens.
And now the most interesting case: the plot with the decimals. For this case each of the stem values represent themselves exactly how they are (if its a 3, it means 31=3), but, the leaves represent their decimals. For a stem and leaf plot with decimals the line in between the stem and leaf column represents the decimal point.
Now that you know how to read a stem and leaf plot, let us construct one from scratch!
How to construct a stem and leaf plot
Let us learn how to create a stem and leaf plot using the next scenario:
Imagine you want to learn the ages of the people who attend the same class of yoga that you attend in the afternoon. After asking all 25 of them, you obtain the following numbers as their ages: 21, 16, 34, 33, 57, 18, 44, 41, 63, 72, 54, 44, 39, 30, 45, 45, 61, 18, 29, 27, 55, 48, 59, 66, 70.
We start by ordering the numbers:
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
Then, we identify the tens that will be part of the stem column (in this case, the first digit of each number), and separate the data values accordingly:
Figure 4: Separating the values according to the stem they belong to
With this in mind, construct the stem and leaf plot by using the first digits of each value as stems and the remaining digits as leaves:
Figure 5: Constructing a stem and leaf plot
And so, the steps you need to follow in order to make a stem and leaf plot are:
Order the data values provided from smallest to largest
Identify the tens or multiples of ten you want to use as stems; remember, these do not need to be a single digit, pick whichever digits are most convenient depending the case.
Separate all the data values classifying them according to the stem they belong to.
Build the stem and leaf plot by using the selected stems and writing down as leaves the digits that remain for each data value.
To solidify this process, here is another example for you to work through:
A 100m swimming race took place and the time it took the competing swimmers to finish was recorded:
Figure 6: Table of data values for the finish time of swimmers in a race
Create a stem and leaf plot
Following the steps we learnt above on how to make a stem and leaf plot, we start by ordering the data values and separating them into categories depending on the stem they belong to:
Building the stem and leaf plot:
Figure 8: Stem and leaf plot of race times
Stem and leaf plot advantages
In our past two lessons we talked about the frequency distribution and histograms, and frequency polygons; we learnt these are methods to represent collected data graphically in order for the data to be understood quickly and easily. Although all the options seen on these two lessons (frequency distribution tables, histograms, and frequency polygons) are simple and reliable for this purpose, all of them have a great disadvantage when the class intervals happen to be a range of values rather than an specific outcome: the frequency of an specific outcome value is lost.
Let us exemplify this disadvantage using example 1 from the section above:
So we have the ages of the people in your yoga class, which are: 16, 18, 18, 21, 27, 29, 30, 33, 34, 39, 41, 44, 44, 45, 45, 48, 54, 55, 57, 59, 61, 63, 66, 70, 72.
Next we input the numbers into a frequency distribution table, a histogram and a frequency polygon.
Figure 9: Data from example 1 into different graphic representations
Notice how, although we can see the pattern of the data being displayed in the different ways presented in figure 9, we know about the data by their ranges defined in the class intervals, but we lose sight of each individual data value (our outcome). In other words, we know there are three data values between the range of 10 and 19 years old, but we do not know the specific age; this particular disadvantage can be seen in all three representations in figure 9: we know how many peoples ages are in each range, but we lose the specific value of everyones ages, this does not happen with a stem and leaf plot.
Let us take a look at the stem and leaf plot for example 1 one more time, and now, instead of thinking on it as a table, let us look at it from a graphic perspective:
Figure 10: Comparing the Stem and Leaf plot with a histogram
As you can see, the stem and leaf plot could very well serve as a rotated histogram, with the great advantage that you continue to know each individual data value from the plot while is represented in an efficient way.
How to use a stem and leaf plot
On this section we will take a look at a few other stem and leaf plot examples so you can see how we can use these kind of plots in many ways!
A Stem and Leaf Plot representing the ages of people being immunized against a certain infection.
Figure 11: Stem and leaf plot of peoples ages
Write out a list of all the numerical data included in the plot:
If we count all of the leaves in the right hand side of the plot, we know there are 18 data values in total, 6 for each range: 0-9, 10-19 and 20-29. So, the data values from smallest to biggest are:
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
\;
Sandras class brought all their animals into school one day. The weight of each animal was measured and then displayed in the following Stem and Leaf Plot:
Figure 12: Stem and leaf plot of weight of Sandras animals
What was the combined weight of every single animal that was taken into school that day?
There are 14 weight values in the plot, adding them all together the result is:
combined \, weight = 1+1+4+6+7+7+8+10+11+13+15+22+25+28=158
Equation 1: The combined weight of all Sandras animals
The combined weight is 158 pounds
What was the range of animal weights?
The weight of Sandras animals goes from one pound to 28 pounds.
So the range is largest weight - smallest weight = 28 - 1= 27 pounds
On these last two problem examples we worked on interpreting stem and leaf plots, you can see another interesting example in our last video for this lesson where we introduce what is called the back to back stem and leaf plot (also called a double stem and leaf plot), that video should reinforce what you learned today on how to make a stem and leaf plot and how it can be used.
And so, we are at the end of our lesson for this topic; as always, we have a few link recommendations: On this article you can read more about what are stem and leaf plots used for, and this website showcases how to build a stem and leaf plots step by step. Use these recommendations to supplement your independent studies.
This is it for today, see you in our next lesson!
How to display a data set in a Stem and Leaf Plot
Finding the mean, median, mode and range for a stem and leaf plot
A Stem and Leaf Plot is a good way to represent data. The data is split into a "stem" which represents the first digit, and a "leaf" which represents the last digit.
Another way to understand these parts are that:
The stem contains all digits in the tens place and up
The leaf is the digit in the ones place only
Write out a list of all the numerical data
Sandra's class brought all their animals into school one day. The weight of each animal was measured and then displayed in the following Stem and Leaf Plot:
Weight of Animal in Pounds
Creating a Stem and Leaf Plot
Sometimes when we have too much data lumped into a category we like to split each stem into two or more categories. This sometimes makes the data more nicely represented. For the above question split each stem into two equal parts.
It is often useful to consider two data sets to each other. A "Back-to-Back Stem and Leaf Plot" is a good way of handily comparing two distributions.
Meghan's basketball team has played two seasons so far. The amount of points her team scored in each game is given in the following back-to-back stem and leaf plot:
Amount of Points Scored
List the amount of points her team scored for all their games for each season.
Recognizing the leaf and stem of numbers
Underline the leaf of the number
Circle the stem of the number
Write a number with:
a stem of 0
a leaf of 5
a stem of 123.
Circle the pairs of numbers with the same stem
Which number would have the same number of digits?
Two numbers with the same stem
Two numbers with the same leaf
Ordering numbers using a stem and leaf plot
Sort the data list using a stem and leaf plot
What are the stems in the data set? Write them in order from least to greatest in the stem column of the plot.
Write the leaves for the numbers into the stem and leaf plot.
Order the leaves (digits) in each row from least to greatest
Use the stem and leaf plot you have completed to list all the data values from least to greatest.
What is the range for the data in this stem and leaf plot?
What is the median for the data in this stem and leaf plot?
What is the mode for the data in this stem and leaf plot?
What is the mean for the data in this stem and leaf plot? Give the answer with 2 decimal places.
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RuneScape:Grand Exchange Market Watch/Food Index - The RuneScape Wiki
GEMW/Food
The Food Index is made up of a weighted average of all of the current foods listed in the Market Watch, with the starting date of this average on 12 January 2008, at an index of 100. The overall rising and falling of food prices is reflected in this index.
While specialised for just watching food prices, it is set up and adjusted in a manner similar to the Common Trade Index, and the divisor may be adjusted to include new food added by Jagex (see the FAQ for more information).
Start date: 12 January 2008 (at index of 100)
Number of items: 19 (last adjusted on 12 January 2008)
Index divisor: 19.0000 (last adjusted on 12 January 2008)
107 160 10,000 view 7 hours ago
92 138 10,000 view 7 hours ago
Great white shark 1,402
9 14 1,000 view 7 hours ago
20 30 1,000 view 7 hours ago
80 120 1,000 view 7 hours ago
{\displaystyle {div}_{\text{old}}=19.0000}
{\displaystyle {div}_{\text{new}}={div}_{\text{old}}\times {\frac {\sum \left({\frac {p}{q}}\right)_{\text{new}}}{\sum \left({\frac {p}{q}}\right)_{\text{old}}}}}
{\displaystyle {\begin{aligned}\sum \left({\frac {p}{q}}\right)_{\text{old}}&={\text{sum of ratios prior to change}}\\&={\text{sum of unchanged ratios}}+{\text{sum of removed ratios}}\\&=\left({\frac {196}{175}}+{\frac {225}{195}}+\dots +{\frac {421}{150}}\right)+\left({\frac {810}{887}}+{\frac {276}{32}}+\dots +{\frac {1,102}{100}}\right)\\&=42.27255853{\text{ (up to 8 d.p.)}}\end{aligned}}}
{\displaystyle {\begin{aligned}\sum \left({\frac {p}{q}}\right)_{\text{new}}&={\text{sum of ratios prior to change}}-{\text{sum of removed ratios}}+{\text{sum of added ratios}}\\&=\sum \left({\frac {p}{q}}\right)_{\text{old}}-{\text{sum of removed ratios}}+{\text{number of added items}}\\&=42.27255853-\left({\frac {810}{887}}+{\frac {276}{32}}+\dots +{\frac {1,102}{100}}\right)+7\\&=19.49428715{\text{ (up to 8 d.p.)}}\end{aligned}}}
{\displaystyle {\begin{aligned}{div}_{\text{new}}&={div}_{\text{old}}\times {\frac {\sum \left({\frac {p}{q}}\right)_{\text{new}}}{\sum \left({\frac {p}{q}}\right)_{\text{old}}}}\\&=19.0000\times {\frac {42.27255853}{19.49428715}}\\&=8.7619{\text{ (4 d.p.)}}\end{aligned}}}
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Automorphism_group Knowpia
In mathematics, the automorphism group of an object X is the group consisting of automorphisms of X. For example, if X is a finite-dimensional vector space, then the automorphism group of X is the group of invertible linear transformations from X to itself (the general linear group of X). If instead X is a group, then its automorphism group
{\displaystyle \operatorname {Aut} (X)}
is the group consisting of all group automorphisms of X.
The automorphism group of a field extension
{\displaystyle L/K}
is the group consisting of field automorphisms of L that fix K. If the field extension is Galois, the automorphism group is called the Galois group of the field extension.
The automorphism group of the projective n-space over a field k is the projective linear group
{\displaystyle \operatorname {PGL} _{n}(k).}
The automorphism group
{\displaystyle G}
of a finite cyclic group of order n is isomorphic to
{\displaystyle (\mathbb {Z} /n\mathbb {Z} )^{\times }}
with the isomorphism given by
{\displaystyle {\overline {a}}\mapsto \sigma _{a}\in G,\,\sigma _{a}(x)=x^{a}}
.[2] In particular,
{\displaystyle G}
The automorphism group of a finite-dimensional real Lie algebra
{\displaystyle {\mathfrak {g}}}
has the structure of a (real) Lie group (in fact, it is even a linear algebraic group: see below). If G is a Lie group with Lie algebra
{\displaystyle {\mathfrak {g}}}
, then the automorphism group of G has a structure of a Lie group induced from that on the automorphism group of
{\displaystyle {\mathfrak {g}}}
.[3][4][a]
If G is a group acting on a set X, the action amounts to a group homomorphism from G to the automorphism group of X amounts to a group action on X. Indeed, each left G-action on a set X determines
{\displaystyle G\to \operatorname {Aut} (X),\,g\mapsto \sigma _{g},\,\sigma _{g}(x)=g\cdot x}
, and, conversely, each homomorphism
{\displaystyle \varphi :G\to \operatorname {Aut} (X)}
defines an action by
{\displaystyle g\cdot x=\varphi (g)x}
. This extends to the case when the set X has more structure than just a set. For example, if X is a vector space, then a group action of G on X is a group representation of the group G, representing G as a group of linear transformations (automorphisms) of X; these representations are the main object of study in the field of representation theory.
Here are some other facts about automorphism groups:
{\displaystyle A,B}
be two finite sets of the same cardinality and
{\displaystyle \operatorname {Iso} (A,B)}
the set of all bijections
{\displaystyle A\mathrel {\overset {\sim }{\to }} B}
{\displaystyle \operatorname {Aut} (B)}
, which is a symmetric group (see above), acts on
{\displaystyle \operatorname {Iso} (A,B)}
from the left freely and transitively; that is to say,
{\displaystyle \operatorname {Iso} (A,B)}
is a torsor for
{\displaystyle \operatorname {Aut} (B)}
(cf. #In category theory).
Let P be a finitely generated projective module over a ring R. Then there is an embedding
{\displaystyle \operatorname {Aut} (P)\hookrightarrow \operatorname {GL} _{n}(R)}
, unique up to inner automorphisms.[5]
Automorphism groups appear very naturally in category theory.
{\displaystyle A,B}
are objects in some category, then the set
{\displaystyle \operatorname {Iso} (A,B)}
{\displaystyle A\mathrel {\overset {\sim }{\to }} B}
is a left
{\displaystyle \operatorname {Aut} (B)}
-torsor. In practical terms, this says that a different choice of a base point of
{\displaystyle \operatorname {Iso} (A,B)}
differs unambiguously by an element of
{\displaystyle \operatorname {Aut} (B)}
, or that each choice of a base point is precisely a choice of a trivialization of the torsor.
{\displaystyle X_{1}}
{\displaystyle X_{2}}
are objects in categories
{\displaystyle C_{1}}
{\displaystyle C_{2}}
{\displaystyle F:C_{1}\to C_{2}}
is a functor mapping
{\displaystyle X_{1}}
{\displaystyle X_{2}}
{\displaystyle F}
induces a group homomorphism
{\displaystyle \operatorname {Aut} (X_{1})\to \operatorname {Aut} (X_{2})}
, as it maps invertible morphisms to invertible morphisms.
In particular, if G is a group viewed as a category with a single object * or, more generally, if G is a groupoid, then each functor
{\displaystyle G\to C}
, C a category, is called an action or a representation of G on the object
{\displaystyle F(*)}
, or the objects
{\displaystyle F(\operatorname {Obj} (G))}
. Those objects are then said to be
{\displaystyle G}
-objects (as they are acted by
{\displaystyle G}
); cf.
{\displaystyle \mathbb {S} }
-object. If
{\displaystyle C}
is a module category like the category of finite-dimensional vector spaces, then
{\displaystyle G}
-objects are also called
{\displaystyle G}
Automorphism group functorEdit
{\displaystyle M}
be a finite-dimensional vector space over a field k that is equipped with some algebraic structure (that is, M is a finite-dimensional algebra over k). It can be, for example, an associative algebra or a Lie algebra.
Now, consider k-linear maps
{\displaystyle M\to M}
that preserve the algebraic structure: they form a vector subspace
{\displaystyle \operatorname {End} _{\text{alg}}(M)}
{\displaystyle \operatorname {End} (M)}
. The unit group of
{\displaystyle \operatorname {End} _{\text{alg}}(M)}
is the automorphism group
{\displaystyle \operatorname {Aut} (M)}
. When a basis on M is chosen,
{\displaystyle \operatorname {End} (M)}
is the space of square matrices and
{\displaystyle \operatorname {End} _{\text{alg}}(M)}
is the zero set of some polynomial equations, and the invertibility is again described by polynomials. Hence,
{\displaystyle \operatorname {Aut} (M)}
is a linear algebraic group over k.
Now base extensions applied to the above discussion determines a functor:[6] namely, for each commutative ring R over k, consider the R-linear maps
{\displaystyle M\otimes R\to M\otimes R}
preserving the algebraic structure: denote it by
{\displaystyle \operatorname {End} _{\text{alg}}(M\otimes R)}
. Then the unit group of the matrix ring
{\displaystyle \operatorname {End} _{\text{alg}}(M\otimes R)}
over R is the automorphism group
{\displaystyle \operatorname {Aut} (M\otimes R)}
{\displaystyle R\mapsto \operatorname {Aut} (M\otimes R)}
is a group functor: a functor from the category of commutative rings over k to the category of groups. Even better, it is represented by a scheme (since the automorphism groups are defined by polynomials): this scheme is called the automorphism group scheme and is denoted by
{\displaystyle \operatorname {Aut} (M)}
Level structure, a technique to remove an automorphism group
^ First, if G is simply connected, the automorphism group of G is that of
{\displaystyle {\mathfrak {g}}}
. Second, every connected Lie group is of the form
{\displaystyle {\widetilde {G}}/C}
{\displaystyle {\widetilde {G}}}
is a simply connected Lie group and C is a central subgroup and the automorphism group of G is the automorphism group of
{\displaystyle G}
that preserves C. Third, by convention, a Lie group is second countable and has at most coutably many connected components; thus, the general case reduces to the connected case.
^ Hartshorne 1977, Ch. II, Example 7.1.1.
^ Dummit & Foote 2004, § 2.3. Exercise 26.
^ Hochschild, G. (1952). "The Automorphism Group of a Lie Group". Transactions of the American Mathematical Society. 72 (2): 209–216. JSTOR 1990752.
^ Fulton & Harris 1991, Exercise 8.28.
^ Milnor 1971, Lemma 3.2.
^ Waterhouse 2012, § 7.6.
Milnor, John Willard (1971). Introduction to algebraic K-theory. Annals of Mathematics Studies. Vol. 72. Princeton, NJ: Princeton University Press. ISBN 9780691081014. MR 0349811. Zbl 0237.18005.
Waterhouse, William C. (2012) [1979]. Introduction to Affine Group Schemes. Graduate Texts in Mathematics. Vol. 66. Springer Verlag. ISBN 9781461262176.
https://mathoverflow.net/questions/55042/automorphism-group-of-a-scheme
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RuneScape:Grand Exchange Market Watch/Herb Index - The RuneScape Wiki
Herbdex
GEMW/Herbs
The Herb Index is made up of a weighted average of all of the grimy and clean herbs listed in the Market Watch, with the starting date of this average on 10 June 2009, at an index of 100. The overall rising and falling of herb prices is reflected in this index.
While specialised for just watching herb prices, it is set up and adjusted in a manner similar to the Common Trade Index, and the divisor may be adjusted to include new herbs added by Jagex (see the FAQ for more information).
As of today, this index is 525.12 -4.47
Historical chart[edit source]
Start date: 10 June 2009 (at index of 100)
Change today: -4.47
Number of items: 34 (last adjusted on 6 November 2011)
Index divisor: 33.2409 (last adjusted on 6 November 2011)
List of items[edit source]
This is the current list of items included in this index:
12 18 10,000 view 6 hours ago
1 1 10,000 view 6 hours ago
Grimy spirit weed 9,859
Clean guam 502
Clean marrentill 5,022
Clean tarromin 194
Clean harralander 1,561
8 12 10,000 view 6 hours ago
Clean toadflax 5,093
Clean irit 831
Clean wergali 11,530
Clean spirit weed 9,721
Clean avantoe 1,648
Clean cadantine 2,609
Clean lantadyme 11,870
Clean dwarf weed 14,089
Clean torstol 16,868
Clean fellstalk 12,534
Adjustments[edit source]
{\displaystyle {div}_{\text{old}}=32.0000}
{\displaystyle {div}_{\text{new}}={div}_{\text{old}}\times {\frac {\sum \left({\frac {p}{q}}\right)_{\text{new}}}{\sum \left({\frac {p}{q}}\right)_{\text{old}}}}}
{\displaystyle {\begin{aligned}\sum \left({\frac {p}{q}}\right)_{\text{old}}&={\text{sum of ratios prior to change}}\\&={\frac {121}{490}}+{\frac {25}{192}}+{\frac {121}{216}}+\dots +{\frac {25,400}{2,762}}\\&=51.57722802{\text{ (up to 8 d.p.)}}\end{aligned}}}
{\displaystyle {\begin{aligned}\sum \left({\frac {p}{q}}\right)_{\text{new}}&={\text{sum of ratios prior to change}}-{\text{sum of removed ratios}}+{\text{sum of added ratios}}\\&=\sum \left({\frac {p}{q}}\right)_{\text{old}}-{\text{sum of removed ratios}}+{\text{number of added items}}\\&=51.57722802-0+2\\&=53.57722802{\text{ (up to 8 d.p.)}}\end{aligned}}}
{\displaystyle {\begin{aligned}{div}_{\text{new}}&={div}_{\text{old}}\times {\frac {\sum \left({\frac {p}{q}}\right)_{\text{new}}}{\sum \left({\frac {p}{q}}\right)_{\text{old}}}}\\&=32.0000\times {\frac {53.57722802}{51.57722802}}\\&=33.2409{\text{ (4 d.p.)}}\end{aligned}}}
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Kinetic and mechanistic study of the reaction between methane sulfonamide...
Kinetic and mechanistic study of the reaction between methane sulfonamide (CH 3S(O) 2NH 2) and OH
Berasategui, Matias; Amedro, Damien; Edtbauer, Achim; Williams, Jonathan; Lelieveld, Jos; Crowley, John N.
Methane sulfonamide (MSAM), inline-formulaCH3S(O)2NH2, was recently detected for the first time in ambient air over the Red Sea and the Gulf of Aden where peak mixing ratios of inline-formula≈60 inline-formulapptv were recorded. Prior to this study the rate constant for its reaction with the inline-formulaOH radical and the products thereby formed were unknown, precluding assessment of its role in the atmosphere. We have studied the inline-formulaOH-initiated photo-oxidation of MSAM in air (298 inline-formulaK, 700 inline-formulaTorr total pressure) in a photochemical reactor using in situ detection of MSAM and its products by Fourier transform infrared (FTIR) absorption spectroscopy. The relative rate technique, using three different reference compounds, was used to derive a rate coefficient of inline-formula
M12inlinescrollmathml\left(normal 1.4±normal 0.3\right)×{normal 10}^{-normal 13}\phantom{\rule{0ex}{0ex}}unit{\mathrm{normal cm}}^{normal 3}\phantom{\rule{0ex}{0ex}}\mathrm{normal molec}{.}^{-normal 1}\phantom{\rule{0ex}{0ex}}{\mathrm{normal s}}^{-normal 1}
171pt15ptsvg-formulamathimge77641cad2fb2065a856345613e01084 acp-20-2695-2020-ie00001.svg171pt15ptacp-20-2695-2020-ie00001.png . The main end products of the photo-oxidation observed by FTIR were inline-formulaCO2, inline-formulaCO, inline-formulaSO2, and inline-formulaHNO3 with molar yields of (inline-formula0.73±0.11), (inline-formula0.28±0.04), (inline-formula0.96±0.15), and (inline-formula0.62±0.09), respectively. inline-formulaN2O and inline-formulaHC(O)OH were also observed in smaller yields of (inline-formula0.09±0.02) and (inline-formula0.03±0.01). Both the low rate coefficient and the products formed are consistent with hydrogen abstraction from the inline-formula−CH3 group as the dominant initial step. Based on our results MSAM has an atmospheric lifetime with respect to loss by reaction with inline-formulaOH of about 80 d.
Berasategui, Matias / Amedro, Damien / Edtbauer, Achim / et al: Kinetic and mechanistic study of the reaction between methane sulfonamide (CH3S(O)2NH2) and OH. 2020. Copernicus Publications.
Rechteinhaber: Matias Berasategui et al.
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Linear Light Field Capture - Looking Glass Documentation
Linear Capture Overview
A compressed output of a Linear Capture, rendered in Unreal Engine.
Linear capture of light fields is one of the most common light field capture methods, either in the physical realm with a camera rail, or in digital tools like Blender, Unreal, Unity, 3ds Max, Cinema 4D, and others. If you can move a camera left to right (or right to left), you're set!
Okay, so what are the rules? How much should my camera travel? How far should my subject be from my camera? Great questions! There are a few formulas that will determine where to position your subject and how far your camera will need to travel.
In this example , we have a camera with an FOV (field of view) of 58 degrees (equivalent to a 39mm lens). We can see through the visualization that there's a specific area where the two camera views overlap and some areas at the edges where they don't. So, what's the deal here? In short, as long as your subject is within the overlapping view cone and is visible in all of your views, you'll be able to get a great light field.
The distance between these two triangles is labeled as the travel and the height of the intersection is labeled as the distance. The triangle we want to solve for is actually the white triangle in the middle there, but split in 2 to give us a right angle.
We've got this triangle now, but what does this mean? Will my subject have to be at the focal point between these two views? Not exactly! What we've just determined is the minimum distance in which your subject can be relative to your camera. For a proper photoset, you'll want to ensure that your subject is in the purple region, meaning that it's visible in every view in between!
You can see in the image set above that 48 different triangles are overlapped on top of one another. This is representative of the captures you'll make, where each triangle represents 1 view. You can see that while the inner triangles overlap at different points with one another, the only point where all views overlap is still the original triangle we uncovered in our first steps above.
Subject Size, Camera Distance, and Travel Distance
In general, the closer your subject is to the tip of the triangle, the closer it is to the maximum depth the Looking Glass can provide at maximum focus. This means that in HoloPlay Studio, you'll have to slide the focus slider all the way to the maximum in order to get your object in focus. In practice, this isn't great, since it gives you less freedom overall with the image.
Instead, what you'll want to do is position your subject a bit further away from the inflection point of all these views. This will give you more freedom overall with the values you can set focus to within HoloPlay Studio.
Rules and Mathematics
This formula links the distance of travel to the minimum distance of the camera to the subject. This is dependent on the field of view of your camera. In this example, we're using a camera with a field of view of 58 degrees, or a 39mm lens, for those more photographically inclined.
\tan(camera FOV/2) = (Travel/2)/distance
Whoa, there! That's a bunch of math, what's going on?! Don't worry about this formula too much, it's actually rather simple! In order to determine the distance to the subject, we'll need to plug in our camera's FOV, then this formula will give us the relationship between travel and distance.
Let's say we're using a camera with a 58 degree FOV. In this case, our formula would become:
\tan(58/2) = (Travel/2)/distance
\tan(29) = (Travel/2)/distance
.55 = (Travel/2)/distance
distance*.55 = Travel/2
distance * 1.1 = Travel
If you have any questions about light field capture you can reach out to the team on Discord or post feedback or requests at our feedback forum!
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Compatibility of t-structures for quantum symplectic resolutions
15 September 2016 Compatibility of
t
-structures for quantum symplectic resolutions
Kevin McGerty, Thomas Nevins
W
be a smooth complex variety with the action of a connected reductive group
G
. Adapting Teleman’s stratification approach to a microlocal context, we prove a vanishing theorem for the functor of
G
-invariant sections—that is, of quantum Hamiltonian reduction—for
G
-equivariant twisted
\mathcal{D}
W
. As a consequence, when
W
is affine we establish a sufficient combinatorial condition for exactness of the global sections functors of microlocalization theory. When combined with the derived equivalence results of our recent work, this gives precise criteria for “microlocalization of representation categories” in the spirit of Kashiwara–Rouquier and many other authors.
Kevin McGerty. Thomas Nevins. "Compatibility of
t
-structures for quantum symplectic resolutions." Duke Math. J. 165 (13) 2529 - 2585, 15 September 2016. https://doi.org/10.1215/00127094-3619684
Received: 21 March 2014; Revised: 16 October 2015; Published: 15 September 2016
Secondary: 16S38 , 17B63 , 53D20 , 53D55
Keywords: $D$-modules , $t$-exactness , Kirwan–Ness stratification , Localization , microlocalization , quantum Hamiltonian reduction
Kevin McGerty, Thomas Nevins "Compatibility of
t
-structures for quantum symplectic resolutions," Duke Mathematical Journal, Duke Math. J. 165(13), 2529-2585, (15 September 2016)
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Effects of Trenched Holes on Film Cooling of a Contoured Endwall Nozzle Vane | J. Turbomach. | ASME Digital Collection
e-mail: silvia.ravelli@unibg.it
Barigozzi, G., Franchini, G., Perdichizzi, A., and Ravelli, S. (July 20, 2011). "Effects of Trenched Holes on Film Cooling of a Contoured Endwall Nozzle Vane." ASME. J. Turbomach. July 2012; 134(4): 041009. https://doi.org/10.1115/1.4003658
The present paper investigates the effects of the application of trenched holes in the front part of a contoured film cooled endwall. Two trench configurations were tested, changing the trench depth. Tests have been carried out at low speed
(M2is=0.2)
and low inlet turbulence intensity level, with coolant mass flow rate ratio varied within the 0.5–2.5% range. Pressure probe traverses were performed downstream of the vane trailing edge to show the secondary flow field modifications and to evaluate trench additional losses. Endwall distributions of film cooling effectiveness have been obtained by the thermochromic liquid crystal (TLC) technique. For each injection condition, energy loss coefficient and film cooling effectiveness distributions were analyzed and compared with the ones obtained from rows of cylindrical holes. Laterally and area averaged effectiveness as well as pitch and mass averaged kinetic energy loss coefficients were computed to enlighten any change induced by the introduction of trenched holes. A uniform and high thermal coverage was obtained in the region just downstream of the trench, but it quickly decayed because of enforced mixing of coolant with main flow. Compared with the cylindrical hole configuration, trenches are able to provide a higher global cooling effectiveness, but a larger amount of coolant injection is required. The introduction of both trenches is responsible for a secondary thermodynamic loss increase of about 0.7% at low coolant injection rates. Increasing the blowing rates, the additional loss vanishes.
blades, coolants, cooling, film flow, nozzles, pressure, thermodynamics, turbulence
Coolants, Cooling, Film cooling, Flow (Dynamics), Nozzles, Pressure, Turbulence, Cascades (Fluid dynamics), Momentum
Contouring Effects on the Adiabatic Effectiveness Distribution Over a Film Cooled Endwall Cascade
, Graz, Austria.
Experimental Analysis of the Combining Effects of Endwall Contouring and Film Cooling Through Shaped Holes
Proceedings of the Ninth International Symposium on Experimental and Computational Aerothermodynamics of Internal Flows (ISAIF9)
, Gyeongju, Korea.
Bump and Trench Modifications to Film-Cooling Holes at the Vane Endwall Junction
Jet Mixing in a Trench
Film Cooling From a Row of Holes Embedded in Transverse Slots
EN ISO 5167-2:2003(E).
AGARD CP 390 Conference on Heat Transfer and Cooling in Gas Turbines
, Bergen, Norway, May.
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Planet Earth/2e. Other Sources of Energy: Gravity, Tides, and the Geothermal Gradient. - Wikibooks, open books for an open world
Planet Earth/2e. Other Sources of Energy: Gravity, Tides, and the Geothermal Gradient.
2 Lord Kelvin and the First Scientific Estimate for the Age of Earth
3 Earth’s Interior Thermonuclear Energy
4 Gravity, Tides and Energy from Earth’s Inertia
Geothermal Gradient[edit | edit source]
A schematic view of the geothermal gradient of increasing temperature with depth inside the Earth.
The sun may appear to be Earth’s only source of energy, but there are other much deeper sources of energy hidden inside Earth. In the pursuit of natural resources such as coal, iron, gold and silver during the heights of the industrial revolution, mining engineers and geologists took notice of a unique phenomenon as they dug deeper and deeper into the interior of the Earth. The deeper you travel down into an underground mine, the warmer the temperature becomes. Caves and shallow mines near the surface, take on a yearly average temperature making hot summer days feel cool in a cave and cold winter days feel warm, but as one descends deeper and deeper underground, ambient temperatures begin to increase. Of course, the amount of increase in temperature varies depending on the proximity you are to an active volcano or upwelling magma, but in most regions on land, a descend of 1,000 meters underground will increase temperatures between 25 to 30° Celsius. One of the deepest mines in the world is the TauTona Mine in South Africa, which descends to depths of 3,900 meters with ambient temperatures rising between 55 °C (131 °F) and 60 °C (140 °F), rivaling or topping the hottest temperatures ever recorded on Earth’s surface. Scientists pondered where this energy, this heat within the Earth comes from.
Scientists of the 1850s viewed the Earth like a giant iron ball heated to glowing hot temperatures in the blacksmith-like furnace of the sun and slowly cooling down ever since its formation. Such view of a hot Earth, bore its origins to the rise of industrial iron furnaces that dot the cityscapes of the 1850s, suggested that Earth, like poured molten iron was once molten and over its long history has cooled. Suggesting that the observed heat experienced deep underground in mines was the cooling remnant of Earth’s original heat from a time in its ancient past when it was forged from the sun. Scientists term this original interior heat within Earth left over from its formation, Accretionary heat.
Lord Kelvin and the First Scientific Estimate for the Age of Earth[edit | edit source]
As a teenager, William Thomson pondered the possibility of using this geothermal gradient of heat in Earth’s interior as a method to determine the age of the Earth. He imagined the Earth to have cooled into its current solid rock from an original liquid molten state, and that the temperatures on the surface of the Earth had not changed significantly over the course of its history. The temperature gradient was directly related to how long the Earth had been cooling. Before changing his name from William Thomson to Lord Kelvin, he acquired an accurate set of measurements of the Earth’s geothermal gradient from reports of miners in 1862, and returned to the question of the age of the Earth.
Lord Kelvin assumed three initial criteria, first was that Earth was once a molten hot liquid, with a uniform hot temperature, and second that this initial temperature was about 3,900 °C, hot enough to melt all types of rocks. Lord Kelvin also assumed that the temperature on Earth’s surface would be the same throughout its history near 0 °C. Like a hot potato thrown into an icy freezer, the center of the Earth would retain its heat at its core, while the outer edges of the Earth would cool with time. He devised a simple formula:
{\displaystyle {{\text{Age of Earth}}={\frac {(T/G)^{2}}{\pi k}}}}
Where T is equal to the initial temperature, 3,900 °C. G is the geothermal gradient he estimated to about 36 °C/km from those measurements in mines, and k was the thermal diffusivity, or the rate that a material cools down measured in meters per second. While Lord Kelvin had established estimates for T, G, and used the constant π, he still had to determine k the thermal diffusivity. In his lab, he experimented with various materials, heating them up and measuring how quickly heat was conducted through the material, and found a good value to use for the Earth of 0.0000012 meters squared per second. During these experiments of heating various materials and measuring how quickly they cooled down, Lord Kelvin was aided by his assistant a young student named John Perry. It must have been exciting when Lord Kelvin calculated an age of the Earth to around 93 million years, although he gave a broad range in his 1863 paper between 22 to 400 million years. Lord Kelvin’s estimate gave hope to Charles Darwin’s budding theory of evolution, which required a long history for various lifeforms to evolve, but ran counter to a notion that Earth had always existed.
John Perry who idolized his professor, graduated and moved on to a prestigious teaching position in Tokyo, Japan. It was there in 1894 he was struck by a foolish assumption that they had made in trying to estimate the age of the Earth, and it may have occurred to him after eating some hot soup on the streets of Tokyo. In a boiling pot of soup, heat is not dispersed through conduction the transfer of heat energy by simple direct contact, but dispersed through convection, that is the transfer of energy with the motion of matter, and in the case of the Earth, the interior of the planet may have acted like a pot of boiling soup, the liquid bubbling and churning bringing up not only heat to the surface, but also matter. John Perry realized if the heat transfer of the interior of the Earth was like boiling soup, rather than an iron ball, the geothermal gradient would be prolonged far longer near the surface due to the upwelling of fresh liquid magma from below. In a pot of boiling soup, the upper levels will retain higher temperatures because the liquid is mixing and moving as it is heated on the stove.
Convection of heat transfer (boiling water), versus Conduction of heat transfer (the hot handle of pot).
In 1894, John Perry published a paper in Nature, indicating the error in Lord Kelvin’s previous estimate for the age of the Earth. Today, we know from radiometric dating that the Earth is 4.6 billion years old, 50 times longer than Lord Kelvin’s estimate. John Perry explained the discrepancy, but it was another idea that captured Lord Kelvin’s attention. The existence of an interior source of energy within the Earth, thermonuclear energy, that could also claim to keep the Earth’s interior hot.
Earth’s Interior Thermonuclear Energy[edit | edit source]
Marie Skłodowska Curie, the great scientist.
Unlike the sun, Earth lacks enough mass and gravity to trigger nuclear fusion at its core. However, throughout its interior, the Earth contains a significant number of large atoms (larger than iron) that formed during the initial giant supernova explosion that formed the solar system. Some of these large atoms, such as thorium-232 and uranium-238 are radioactive. These elements have been slowly decaying ever since their creation around the time of the initial formation of the sun, solar system and Earth. The decay of these large atoms into smaller atoms is called nuclear fission. During the decay, these larger atoms are broken into smaller atoms, some of which can also decay into even smaller atoms, like the gas radon which decays into lead. The decay of larger atoms into smaller atoms produces radioactivity, a term coined by Marie Skłodowska-Curie. In 1898, she was able to detect electromagnetic radiation emitted from both thorium and uranium, and later she and her husband demonstrated that radioactive substances produce heat. This discovery was confirmed by another female scientist named Fanny Gates, who demonstrated the effects of heat on radioactive materials, while the equally brilliant female scientist discovered that radioactive solid substances produced from the decay of thorium and uranium, further decay to a radioactive gas, called radon.
Harriet Brooks who discovered Radon.
The New Zealand scientist Ernest Rutherford, who wrote the classic book on radioactivity.
These scientists worked and corresponded closely with a New Zealander, named Ernest Rutherford, who in 1905 published a definitive book on “Radio-activity.” This collection of knowledge begun to tear down the assumptions made by Lord Kelvin. It also introduced a major quandary in Earth sciences. How much of Earth’s interior heat is a product of accretionary heat and how much is a product of thermonuclear heat from the decay of thorium and uranium?
A century of technology has resulted in breakthroughs in measuring nuclear decay within the interior of the Earth. Nuclear fusion in the sun causes beta plus (β+) decay, in which a proton is converted to a neutron, and generates a positron and neutrino, as well as electromagnetic radiation. In nuclear fission, in which atoms break apart, beta minus (β−) decay occurs. Beta minus (β−) decay causes a neutron to convert to a proton, and generates an electron and antineutrino as well as electromagnetic radiation. If a positron comes in contact with an electron the two sub-atomic particles annihilate each other. If a neutrino comes in contact with an antineutrino the two sub-atomic particles annihilate each other. Most positrons are annihilated in the upper regions of the sun, which are enriched in electrons, while neutrinos are free to blast across space, zipping unseen through the Earth, and are only annihilated if they come in contact with antineutrinos produced by radioactive beta minus (β−) decay from nuclear fission on Earth.
Any time of day, trillions of neutrinos are zipping through your body, followed by a few antineutrinos produced by background radiation. Neither of these subatomic particles cause any health concerns, as they cannot break atomic bonds. However, if they strike a proton, they can emit a tiny amount of energy, in the form of a nearly instantaneous flash of electromagnetic radiation.
The Kamioka Liquid-scintillator Anti-Neutrino Detector in Japan is a complex experiment designed to detect anti-neutrinos emitted during radioactive beta minus (β−) decay caused by both nuclear reactors in energy generating power plants, as well as natural background radiation from thorium-232 and uranium-238 inside the Earth.
The detector is buried deep in an old mine, and consists of a steel sphere filled with a balloon filled liquid scintillator, and buffered by a layer of mineral oil. Light within the steel sphere is detected by highly sensitive phototubes mounted on the inside surface of the steel sphere. Inside the pitch-black sphere any tiny flash of electromagnetic radiation can be detected by the thousands of phototubes that line the surface of the sphere. These phototubes record tiny electrical pulses, which result from the collision of antineutrinos striking protons. Depending on the source of the antineutrinos, they will produce differing amounts of energy in the electrical pulses. Antineutrinos produced by nearby nuclear reactors can be detected, as well as natural antineutrinos caused by the fission of thorium-232 and uranium-238. A census of background electrical pulses indicates that Earth’s interior thermonuclear energy accounts for about 25% of the total interior energy of the Earth (2011 Nature Geoscience 4:647–651, but see 2013 calculations at https://arxiv.org/abs/1303.4667) the other 75% is the accretionary heat, left over from the initial formation of the Earth. Thorium-232 is more abundant near the core of the Earth, while uranium-238 is found closer to the surface. Both elements contribute to enhancing the geothermal gradient observed in Earth’s interior, and extending Earth’s interior energy beyond that predicated for a model involving a cooling Earth with only heat left over from its formation. A few other radioactive elements contribute to Earth’s interior heat, such as potassium-40, but the majority of Earth’s interior energy is a result of residual heat from its formation.
Comparing the total amount of Earth’s interior energy sources with the amount Earth receives via the Sun, reveals an order of magnitude of difference. The entire interior energy from Earth accounts for only about 0.03% of Earth’s total energy. The other 99.97% comes from the sun’s energy, as measured above the atmosphere. It is important to note that it is estimated that current human populations utilize about 30 Tetrawatts or about 0.02% of Earth’s total energy. Hence, the interior energy of Earth and the resulting geothermal gradient could support much of the energy demands of large populations of humans, despite the fact that it accounts for a small amount of Earth’s total energy budget.
Gravity, Tides and Energy from Earth’s Inertia[edit | edit source]
While the vast amount of Earth’s energy comes from the Sun, and a small amount comes from the interior of the Earth, a complete census of Earth’s energy should also discuss a tiny component of Earth’s energy that is derived from its motion and the oscillations of its gravitational pull with both the Moon and the Sun.
Animation of tides as the Moon goes round the Earth with the Sun on the right.
Ocean and Earth tides are caused by the joint gravitational pull of the Moon and Sun. They daily cycle between high and low tides over a longer two-week period. Twice a lunar month, around the new moon and full moon, when a straight line can be draw through the center of the Sun, Moon and Earth, a configuration known as a syzygy, the tidal force of the Moon is reinforced by the gravitational force of the Sun, resulting in a higher than usual tides called a spring tide. When a line drawn through the center of the Sun to the Earth, and Moon to the Earth forms a 90° angle, or is perpendicular, the gravitational force of the Sun partially cancels the gravitational force of the moon, resulting in a weakened tide, called a neap tide. These occur when the Moon is at first quarter or third quarter in the night sky.
Gravitational pull of the Moon generates a tide-generating force, effecting both liquid water, as well as the solid interior of Earth.
Daily tides are a result of Earth’s rotation relative to the position of the Moon. Tides can affect both the solid interior of the Earth (Earth tides), as well as the liquid ocean waters (Ocean tides), which are more noticeable, as ocean waters rise and fall along coastlines. Long records of sea level are averaged to indicate the average sea level along the coastline. The highest astronomical tide and lowest astronomical tide are also recorded, with the lowest record of the tide equivalent on navigational charts as the datum. Metrological conditions (such as hurricanes), as well as tsunamis (caused by earthquakes) can dramatically rise or lower sea level alongs coasts, well beyond the highest and lowest astronomical tides. It is estimated that tides contribute only 3.7 Tetrawatts of energy (Global Climate and Energy Project, Hermann, 2006 Energy), or about 0.002% of Earth’s total energy.
In this census of Earth’s energy, we did not include wind and fossil fuels such as coal, oil and natural gas, as these sources of energy are ultimately a result of input of solar irradiation. Wind is a result of thermal and pressure gradients in the atmosphere, that you will learn more of later when you read about the atmosphere, while fossil fuels are stored biological energy, due to sequestration of organic matter produced by photosynthesis, in the form of hydrocarbons, that you will learn more of as you read about life in a later chapter.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Planet_Earth/2e._Other_Sources_of_Energy:_Gravity,_Tides,_and_the_Geothermal_Gradient.&oldid=3967420"
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Difference between revisions of "Documentation Talk:Reference Section 4" - POV-Wiki
Difference between revisions of "Documentation Talk:Reference Section 4"
m (→inline pictures rendered from the portfolio)
m (Reverted edits by Valera (Talk) to last revision by Jholsenback)
: The issue of new images is something that's been raised before (other sections) ... it's on my list but low priority at the moment --[[User:Jholsenback|jholsenback]] 17:30, 24 October 2010 (UTC)
[http://www.superiorpapers.com/ custom papers]
{\displaystyle {density}={strength}\cdot \left(1-\left({\frac {\min(distance,radius)}{radius}}\right)^{2}\right)^{2}}
{\displaystyle {\textit {density}}={\textit {strength}}\cdot \left(1-\left({\frac {\min({\textit {distance}},{\textit {radius}})}{\textit {radius}}}\right)^{2}\right)^{2}}
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The book-to-market ratio is one indicator of a company's value. The ratio compares a firm's book value to its market value. A company's book value is calculated by looking at the company's historical cost, or accounting value. A firm's market value is determined by its share price in the stock market and the number of shares it has outstanding, which is its market capitalization.
The book-to-market ratio helps investors find a company's value by comparing the firm's book value to its market value.
A high book-to-market ratio might mean that the market is valuing the company's equity cheaply compared to its book value.
The book-to-market ratio compares a company's book value to its market value. The book value is the value of assets minus the value of the liabilities. The market value of a company is the market price of one of its shares multiplied by the number of shares outstanding. The book-to-market ratio is a useful indicator for investors who need to assess the value of a company.
The formula for the book-to-market ratio is the following:
\text{Book-to-Market}=\frac{\text{Common Shareholders' Equity}}{\text{Market Cap}}
Book-to-Market=Market CapCommon Shareholders’ Equity
What Does the Book-to-Market Ratio Tell You?
If the market value of a company is trading higher than its book value per share, it is considered to be overvalued. If the book value is higher than the market value, analysts consider the company to be undervalued. The book-to-market ratio is used to compare a company’s net asset value or book value to its current or market value.
The book value of a firm is its historical cost or accounting value calculated from the company’s balance sheet. Book value can be calculated by subtracting total liabilities, preferred shares, and intangible assets from the total assets of a company. In effect, the book value represents how much a company would have left in assets if it went out of business today. Some analysts use the total shareholders' equity figure on the balance sheet as the book value.
The market value of a publicly-traded company is determined by calculating its market capitalization, which is simply the total number of shares outstanding multiplied by the current share price. The market value is the price that investors are willing to pay to acquire or sell the stock in the secondary markets. Since it is determined by supply and demand in the market, it does not always represent the actual value of a firm.
How to Use the Book-to-Market Ratio
The book-to-market ratio identifies undervalued or overvalued securities by taking the book value and dividing it by the market value. The ratio determines the market value of a company relative to its actual worth. Investors and analysts use this comparison ratio to differentiate between the true value of a publicly-traded company and investor speculation.
In basic terms, if the ratio is above 1, then the stock is undervalued. If it is less than 1, the stock is considered overvalued. A ratio above 1 indicates that the stock price of a company is trading for less than the worth of its assets. A high ratio is preferred by value managers who interpret it to mean that the company is a value stock—that is, it is trading cheaply in the market compared to its book value.
A book-to-market ratio below 1 implies that investors are willing to pay more for a company than its net assets are worth. This could indicate that the company has healthy future profit projections and investors are willing to pay a premium for that possibility. Technology companies and other companies in industries that do not have a lot of physical assets tend to have a low book-to-market ratio.
Difference Between the Book-to-Market Ratio and Market-to-Book Ratio
The market-to-book ratio, also called the price-to-book ratio, is the reverse of the book-to-market ratio. Like the book-to-market ratio, it seeks to evaluate whether a company's stock is over or undervalued by comparing the market price of all outstanding shares with the net assets of the company.
A market-to-book ratio above 1 means that the company’s stock is overvalued. A ratio below 1 indicates that it may be undervalued; the reverse is the case for the book-to-market ratio. Analysts can use either ratio to run a comparison on the book and market value of a firm.
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This problem is a checkpoint for writing equations for arithmetic and geometric sequences. It will be referred to as Checkpoint 5B.
t(n)=1,4,7,10,\ \dots
t(n)=3,\frac{3}{2},\frac{3}{4},\frac{3}{8},\dots
n→t(n)
\begin{array}{c|c} \quad n\quad & \; t(n) \;\\ \hline 1 & 17\\ \hline 2 & \\ \hline 3 & 3\\ \hline 4 & \end{array}
\begin{array}{c|c} \quad n\quad & \;\; t(n) \;\; \\ \hline 1 & \\ \hline 2 & 7.2\\ \hline 3 & 8.64\\ \hline 4 & \end{array}
t(7)=1056
t(12)=116
t(4)
If you have an eBook for CCG, login and then click the following link: Checkpoint 5B: Writing Equations for Arithmetic and Geometric Sequences
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Jeynysha has a Shape Bucket with a trapezoid, right triangle, scalene triangle, parallelogram, square, rhombus, pentagon, and kite. If she reaches in the bucket and randomly selects a shape, find:
P(\text{at least one pair of parallel sides})
Which shapes have at least one pair of parallel sides?
\frac{1}{2}
P(\text{hexagon})
Do any shapes have six sides?
P(\text{not a triangle})
\frac{3}{4}
P(\text{has at least } 3\text{ sides})
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Writing an exam paper with Julia and Jupyter Lab -- Throw All Things Away
Writing an exam paper with Julia and Jupyter Lab
I just taught Linear Algebra, which means I need to give exams with a lot of questions about computations. I would like to have the numbers in this problems to be random, and I like to solve them with computers. (Sorry, students. Only you need to compute by hand. 😆)
I have tried to use Mathematica to do so. The math parts works well. But the typesetting of Mathematica looks quite ugly.
Eventually, I switched to Julia for computations and plotting, and Jupyter Lab for editing the code. I am very happy with the end result.
Here is the Jupyter notebook for one of my exam
In the following, I will tell you how I did it.
Jupyter Lab and IJulia
Jupyter Lab (JL) is an editor which runs in the browser. To use Julia with Jupyter Lab, you need IJulia.
The standard documents you can write with JL is called Jupyter Notebooks, which are very similar to Mathematica notebooks. In other words, you run code chunks called cells in the notebook. The output is then rendered into math equations (if it's LaTeX) or images (if it's a plot) and showed right below the input cell. So you can experiment with your code and immediate see the result (an exam problem in this case). Here is how my exam paper looks like as a Jupyter notebook
It is also very easy to convert Jupyter notebooks into PDF. (Behind the scene, a notebook is first converted into a LaTeX file, and then compiled into PDF.) And here is how my exam paper looks after being converted into PDF.
Julia includes a standard package LinearAlgebra, which does most things a first Linear Algebra course does. However, it uses numeric methods most of the time. If you want symbolic result, you need SymPy and SymPy.jl. The former is a computer algebra system written in Python. The latter makes using SymPy in Julia a bit easier.
You may also want to install RowEchelon.jl. It turns a matrix into its reduced echelon form.
The following code generates a random
3 \times 3
3×3 matrix with integer entries in
\{-1, \ldots, 5\}
{−1,…,5}.
A = rand(-1:5, (3,3))
However, you don't want to get a different matrix each time you run the cell. So at the beginning of each problem, you can seed the random number generator like this.
This guarantees you get the same random matrix each time. And if you don't like the result, just change 1234 to another (fixed) number.
To generate LaTeX programmatically, you need two packages Latexify.jl and LaTeXStrings.jl.
Let's say you want to start your exam problem with something like
A = \cdots
A=⋯ be a
5 \times 5
5×5 matrix.
and you want to replace the
\cdots by the random matrix A which we just generated, you can write
A = %$(latexify(A))
Here the L in front of the string indicates that we want the string to be interpreted as LaTeX code, so that Jupyter Lab knows to render it as a math equation. The %$(latexify(A)) is executed and the result is inserted into the string. The function call latexify(A) turns the Julia object A into a piece of LaTeX representing it. The result will be something like
-1 & 3 & 1 & 18 \\
And Jupyter Lab will display it like this –
A = \left[ \begin{array}{cccc} -1 & 3 & 1 & 18 \\ 1 & 1 & 2 & 8 \\ 1 & 5 & 5 & 34 \\ \end{array} \right]
A=⎣⎢⎡−11131512518834⎦⎥⎤
Defining LaTeX Marcos
Jupyter Lab uses a JavaScript library MathJax to render LaTeX. It works quite well, but certainly it is not exactly the same as LaTeX. For example, if you want to define a LaTeX macro which you can both use in Jupyter Lab, i.e., recognized by MathJax, and also works in LaTeX, which is an intermediate stage when a notebook is converted into PDF, you need to use a trick –
Create a markdown cell and write
\newcommand{\require}[1]{}
$$\require{begingroup}\require{newcommand}$$
\gdef\dsR{{\mathbb{R}}}
Then the macro \dsR will work both in the notebook and in LaTeX.
Raw Cells are a special type of cell in which you can write anything and Jupyter will not try to do anything about it. And when a notebook is converted to LaTeX, whatever in a raw cell will be copied to the LaTeX. So if you want to add some LaTeX code, like a line by the end of the paper, you can create a raw cell at the end of the notebook and write
Good bye! I will see you around.
And this is you will see in the end of the PDF
Note, you can also just write LaTeX in markdown cells, but keep them in raw cells make things a bit cleaner.
Some things which I have not figured out
One thing that bothers me is that I cannot change the preamble of the LaTeX generated from a notebook This a bit annoying because then I cannot control things like the font size of the final PDF, unless I first convert the notebook to LaTeX and change the generated LaTeX file manually. Maybe I will try customized templates next time when I teach Linear Algebra.
Tags: teaching tech LaTeX Julia
© Xing Shi Cai. First published: March 08, 2022. Last modified: May 06, 2022. Website built with Franklin.jl and the Julia programming language. Icons made by Freepik from www.flaticon.com.
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tensor(deprecated)/partial_diff - Maple Help
Home : Support : Online Help : tensor(deprecated)/partial_diff
compute the partial derivatives of a tensor_type with respect to given coordinates
partial_diff( U, coord)
tensor_type whose partial derivatives are to be found
Given the coordinate variables, coord, and any tensor_type U, partial_diff(U, coord) constructs the partial derivatives of U, which will be a new tensor_type (not necessarily a tensor, of course) of rank one higher than that of U.
The extra index due to differentiation is of covariant character, by convention. Thus the index_char field of the result is
[{U}_{\mathrm{index_char}},-1]
Simplification: This routine uses the `tensor/partial_diff/simp` routine for simplification purposes. The simplification routine is applied to each component of result after it is computed. By default, `tensor/partial_diff/simp` is initialized to the `tensor/simp` routine. It is recommended that the `tensor/partial_diff/simp` routine be customized to suit the needs of the particular problem.
When computing the first and second partial derivatives of the covariant metric tensor components, it is suggested that the tensor[d1metric] and tensor[d2metric] routines be used instead of the partial_diff routine so that the symmetries of the first and second partials be implemented using the tensor package indexing functions. The partial_diff routine does not preserve any symmetric properties that the indices of its input may have.
This function is part of the tensor package, and so can be used in the form partial_diff(..) only after performing the command with(tensor) or with(tensor, partial_diff). The function can always be accessed in the long form tensor[partial_diff](..).
\mathrm{with}\left(\mathrm{tensor}\right):
\mathrm{coord}≔[r,\mathrm{\theta },\mathrm{\psi }]:
A≔\mathrm{array}\left(1..3,[f\left(r\right),g\left(\mathrm{\theta }\right),h\left(\mathrm{\psi }\right)]\right):
U≔\mathrm{create}\left([1],\mathrm{op}\left(A\right)\right)
\textcolor[rgb]{0,0,1}{U}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\textcolor[rgb]{0,0,1}{\mathrm{index_char}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{1}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{compts}}\textcolor[rgb]{0,0,1}{=}[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{r}\right)& \textcolor[rgb]{0,0,1}{g}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{\theta }}\right)& \textcolor[rgb]{0,0,1}{h}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{\psi }}\right)\end{array}]]\right)
\mathrm{part_U}≔\mathrm{partial_diff}\left(U,\mathrm{coord}\right)
\textcolor[rgb]{0,0,1}{\mathrm{part_U}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\textcolor[rgb]{0,0,1}{\mathrm{index_char}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-1}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{compts}}\textcolor[rgb]{0,0,1}{=}[\begin{array}{ccc}\frac{\textcolor[rgb]{0,0,1}{ⅆ}}{\textcolor[rgb]{0,0,1}{ⅆ}\textcolor[rgb]{0,0,1}{r}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{r}\right)& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0}& \frac{\textcolor[rgb]{0,0,1}{ⅆ}}{\textcolor[rgb]{0,0,1}{ⅆ}\textcolor[rgb]{0,0,1}{\mathrm{\theta }}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{g}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{\theta }}\right)& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \frac{\textcolor[rgb]{0,0,1}{ⅆ}}{\textcolor[rgb]{0,0,1}{ⅆ}\textcolor[rgb]{0,0,1}{\mathrm{\psi }}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{h}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{\psi }}\right)\end{array}]]\right)
V≔\mathrm{create}\left([],H\left(r,\mathrm{\theta },\mathrm{\psi }\right)\right)
\textcolor[rgb]{0,0,1}{V}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\textcolor[rgb]{0,0,1}{\mathrm{index_char}}\textcolor[rgb]{0,0,1}{=}[]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{compts}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{H}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\theta }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\psi }}\right)]\right)
\mathrm{part_V}≔\mathrm{partial_diff}\left(V,\mathrm{coord}\right)
\textcolor[rgb]{0,0,1}{\mathrm{part_V}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\textcolor[rgb]{0,0,1}{\mathrm{index_char}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{-1}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{compts}}\textcolor[rgb]{0,0,1}{=}[\begin{array}{ccc}\frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{r}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{H}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\theta }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\psi }}\right)& \frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{\mathrm{\theta }}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{H}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\theta }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\psi }}\right)& \frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{\mathrm{\psi }}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{H}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\theta }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\psi }}\right)\end{array}]]\right)
tensor(deprecated)/directional_diff
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Interior-point method - Wikipedia
Algorithms for solving convex optimization problems
Example search for a solution. Blue lines show constraints, red points show iterated solutions.
Interior-point methods (also referred to as barrier methods or IPMs) are a certain class of algorithms that solve linear and nonlinear convex optimization problems.
An interior point method was discovered by Soviet mathematician I. I. Dikin in 1967 and reinvented in the U.S. in the mid-1980s. In 1984, Narendra Karmarkar developed a method for linear programming called Karmarkar's algorithm, which runs in provably polynomial time and is also very efficient in practice. It enabled solutions of linear programming problems that were beyond the capabilities of the simplex method. Contrary to the simplex method, it reaches a best solution by traversing the interior of the feasible region. The method can be generalized to convex programming based on a self-concordant barrier function used to encode the convex set.
Any convex optimization problem can be transformed into minimizing (or maximizing) a linear function over a convex set by converting to the epigraph form.[1] The idea of encoding the feasible set using a barrier and designing barrier methods was studied by Anthony V. Fiacco, Garth P. McCormick, and others in the early 1960s. These ideas were mainly developed for general nonlinear programming, but they were later abandoned due to the presence of more competitive methods for this class of problems (e.g. sequential quadratic programming).
Yurii Nesterov, and Arkadi Nemirovski came up with a special class of such barriers that can be used to encode any convex set. They guarantee that the number of iterations of the algorithm is bounded by a polynomial in the dimension and accuracy of the solution.[2]
Karmarkar's breakthrough revitalized the study of interior-point methods and barrier problems, showing that it was possible to create an algorithm for linear programming characterized by polynomial complexity and, moreover, that was competitive with the simplex method. Already Khachiyan's ellipsoid method was a polynomial-time algorithm; however, it was too slow to be of practical interest.
The class of primal-dual path-following interior-point methods is considered the most successful. Mehrotra's predictor–corrector algorithm provides the basis for most implementations of this class of methods.[3]
1 Primal-dual interior-point method for nonlinear optimization
Primal-dual interior-point method for nonlinear optimization[edit]
The primal-dual method's idea is easy to demonstrate for constrained nonlinear optimization. For simplicity, consider the all-inequality version of a nonlinear optimization problem:
{\displaystyle f(x)}
{\displaystyle c_{i}(x)\geq 0~{\text{for}}~i=1,\ldots ,m,~x\in \mathbb {R} ^{n},}
{\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} ,c_{i}:\mathbb {R} ^{n}\rightarrow \mathbb {R} \quad (1).}
This inequality-constrained optimization problem is then solved by converting it into an unconstrained objective function whose minimum we hope to find efficiently. Specifically, the logarithmic barrier function associated with (1) is
{\displaystyle B(x,\mu )=f(x)-\mu \sum _{i=1}^{m}\log(c_{i}(x)).\quad (2)}
{\displaystyle \mu }
is a small positive scalar, sometimes called the "barrier parameter". As
{\displaystyle \mu }
converges to zero the minimum of
{\displaystyle B(x,\mu )}
should converge to a solution of (1).
The barrier function gradient is
{\displaystyle g_{b}(x,\mu ):=\nabla B(x,\mu )=g(x)-\mu \sum _{i=1}^{m}{\frac {1}{c_{i}(x)}}\nabla c_{i}(x),\quad (3)}
{\displaystyle g(x):=\nabla f(x)}
is the gradient of the original function
{\displaystyle f(x)}
{\displaystyle \nabla c_{i}}
{\displaystyle c_{i}}
In addition to the original ("primal") variable
{\displaystyle x}
we introduce a Lagrange multiplier-inspired dual variable
{\displaystyle \lambda \in \mathbb {R} ^{m}}
{\displaystyle c_{i}(x)\lambda _{i}=\mu ,\forall i=1,\ldots ,m.\quad (4)}
(4) is sometimes called the "perturbed complementarity" condition, for its resemblance to "complementary slackness" in KKT conditions.
We try to find those
{\displaystyle (x_{\mu },\lambda _{\mu })}
for which the gradient of the barrier function is zero.
Applying (4) to (3), we get an equation for the gradient:
{\displaystyle g-A^{T}\lambda =0,\quad (5)}
{\displaystyle A}
is the Jacobian of the constraints
{\displaystyle c(x)}
The intuition behind (5) is that the gradient of
{\displaystyle f(x)}
should lie in the subspace spanned by the constraints' gradients. The "perturbed complementarity" with small
{\displaystyle \mu }
(4) can be understood as the condition that the solution should either lie near the boundary
{\displaystyle c_{i}(x)=0}
, or that the projection of the gradient
{\displaystyle g}
on the constraint component
{\displaystyle c_{i}(x)}
normal should be almost zero.
Applying Newton's method to (4) and (5), we get an equation for
{\displaystyle (x,\lambda )}
{\displaystyle (p_{x},p_{\lambda })}
{\displaystyle {\begin{pmatrix}W&-A^{T}\\\Lambda A&C\end{pmatrix}}{\begin{pmatrix}p_{x}\\p_{\lambda }\end{pmatrix}}={\begin{pmatrix}-g+A^{T}\lambda \\\mu 1-C\lambda \end{pmatrix}},}
{\displaystyle W}
{\displaystyle B(x,\mu )}
{\displaystyle \Lambda }
is a diagonal matrix of
{\displaystyle \lambda }
{\displaystyle C}
is a diagonal matrix with
{\displaystyle C_{ii}=c_{i}(x)}
Because of (1), (4) the condition
{\displaystyle \lambda \geq 0}
should be enforced at each step. This can be done by choosing appropriate
{\displaystyle \alpha }
{\displaystyle (x,\lambda )\to (x+\alpha p_{x},\lambda +\alpha p_{\lambda }).}
Trajectory of the iterates of x by using the interior point method.
^ Boyd, Stephen; Vandenberghe, Lieven (2004). Convex Optimization. Cambridge: Cambridge University Press. p. 143. ISBN 978-0-521-83378-3. MR 2061575.
^ Wright, Margaret H. (2004). "The interior-point revolution in optimization: History, recent developments, and lasting consequences". Bulletin of the American Mathematical Society. 42: 39–57. doi:10.1090/S0273-0979-04-01040-7. MR 2115066.
^ Potra, Florian A.; Stephen J. Wright (2000). "Interior-point methods". Journal of Computational and Applied Mathematics. 124 (1–2): 281–302. doi:10.1016/S0377-0427(00)00433-7.
Dikin, I.I. (1967). "Iterative solution of problems of linear and quadratic programming". Dokl. Akad. Nauk SSSR. 174 (1): 747–748.
Bonnans, J. Frédéric; Gilbert, J. Charles; Lemaréchal, Claude; Sagastizábal, Claudia A. (2006). Numerical optimization: Theoretical and practical aspects. Universitext (Second revised ed. of translation of 1997 French ed.). Berlin: Springer-Verlag. pp. xiv+490. doi:10.1007/978-3-540-35447-5. ISBN 978-3-540-35445-1. MR 2265882.
Karmarkar, N. (1984). "A new polynomial-time algorithm for linear programming" (PDF). Proceedings of the sixteenth annual ACM symposium on Theory of computing – STOC '84. p. 302. doi:10.1145/800057.808695. ISBN 0-89791-133-4. Archived from the original (PDF) on 28 December 2013.
Mehrotra, Sanjay (1992). "On the Implementation of a Primal-Dual Interior Point Method". SIAM Journal on Optimization. 2 (4): 575–601. doi:10.1137/0802028.
Nocedal, Jorge; Stephen Wright (1999). Numerical Optimization. New York, NY: Springer. ISBN 978-0-387-98793-4.
Press, WH; Teukolsky, SA; Vetterling, WT; Flannery, BP (2007). "Section 10.11. Linear Programming: Interior-Point Methods". Numerical Recipes: The Art of Scientific Computing (3rd ed.). New York: Cambridge University Press. ISBN 978-0-521-88068-8.
Wright, Stephen (1997). Primal-Dual Interior-Point Methods. Philadelphia, PA: SIAM. ISBN 978-0-89871-382-4.
Boyd, Stephen; Vandenberghe, Lieven (2004). Convex Optimization (PDF). Cambridge University Press.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Interior-point_method&oldid=1069107991"
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Cross-validated linear regression model for high-dimensional data - MATLAB - MathWorks Italia
Create Cross-Validated Linear Regression Model
Cross-validated linear regression model for high-dimensional data
RegressionPartitionedLinear is a set of linear regression models trained on cross-validated folds. To obtain a cross-validated, linear regression model, use fitrlinear and specify one of the cross-validation options. You can estimate the predictive quality of the model, or how well the linear regression model generalizes, using one or more of these “kfold” methods: kfoldPredict and kfoldLoss.
Unlike other cross-validated, regression models, RegressionPartitionedLinear model objects do not store the predictor data set.
CVMdl = fitrlinear(X,Y,Name,Value) creates a cross-validated, linear regression model when Name is either 'CrossVal', 'CVPartition', 'Holdout', or 'KFold'. For more details, see fitrlinear.
Trained — Linear regression models trained on cross-validation folds
cell array of RegressionLinear model objects
Linear regression models trained on cross-validation folds, specified as a cell array of RegressionLinear models. Trained has k cells, where k is the number of folds.
y={x}_{100}+2{x}_{200}+e.
X=\left\{{x}_{1},...,{x}_{1000}\right\}
Cross-validate a linear regression model. To increase execution speed, transpose the predictor data and specify that the observations are in columns.
CVMdl = fitrlinear(X,Y,'CrossVal','on','ObservationsIn','columns');
CVMdl is a RegressionPartitionedLinear cross-validated model. Because fitrlinear implements 10-fold cross-validation by default, CVMdl.Trained contains a cell vector of ten RegressionLinear models. Each cell contains a linear regression model trained on nine folds, and then tested on the remaining fold.
Predict responses for out-of-fold observations and estimate the generalization error by passing CVMdl to kfoldPredict and kfoldLoss, respectively.
oofYHat = kfoldPredict(CVMdl);
The estimated, generalization, mean squared error is 0.1748.
y={x}_{100}+2{x}_{200}+e.
X=\left\{{x}_{1},...,{x}_{1000}\right\}
1{0}^{-5}
1{0}^{-1}
RegressionLinear | fitrlinear | kfoldPredict | kfoldLoss
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10.7: Conditional Probability - Statistics LibreTexts
https://stats.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fstats.libretexts.org%2FBookshelves%2FIntroductory_Statistics%2FBook%253A_Statistical_Thinking_for_the_21st_Century_(Poldrack)%2F10%253A_Probability%2F10.07%253A_Conditional_Probability
Let’s take the 2016 US Presidential election as an example. There are two simple probabilities that we could use to describe the electorate. First, we know the probability that a voter in the US affiliated with the Republican party:
p(Republican) = 0.44
. We also know the probability that a voter cast their vote in favor of Donald Trump: Trump voter)=0.46. However, let’s say that we want to know the following: What is the probability that a person cast their vote for Donald Trump, given that they are a Republican?
To compute the conditional probability of A given B (which we write as
P(A|B)
, “probability of A, given B”), we need to know the joint probability (that is, the probability of both A and B occurring) as well as the overall probability of B:
P(A|B) = \frac{P(A \cap B)}{P(B)}
Figure 10.3: A graphical depiction of conditional probability, showing how the conditional probability limits our analysis to a subset of the data.
It can be useful to think of this is graphically. Figure 10.3 shows a flow chart depicting how the full population of voters breaks down into Republicans and Democrats, and how the conditional probability (conditioning on party) further breaks down the members of each party according to their vote.
10.7: Conditional Probability is shared under a not declared license and was authored, remixed, and/or curated by Russell A. Poldrack via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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Parallel Computing and Computer Clusters/Theory - Wikibooks, open books for an open world
Parallel Computing and Computer Clusters/Theory
1 Typical real-world applications
2 Classes of problems
2.1 Embarrassingly parallel problems
3 Parallel Programming Models
4 History of distributed computing
4.1 The Eighties and Nineties: PVM and MPI
4.2 Today: GFS, MapReduce, Hadoop
5 Mathematics of parallel processing
Typical real-world applications[edit | edit source]
What do people do with clusters of computers?
Scientific and engineering work
Weather modeling and prediction
Classes of problems[edit | edit source]
Embarrassingly parallel problems[edit | edit source]
Parallel Programming Models[edit | edit source]
There are generally two ways to accomplish parallel architectures. They can be either used separately or the architecture can be any combination of the two. The shared memory model is a model where all processors in the architecture share memory and address spaces. All the processors in the architecture can access all of the attached main memory. The message passing model is a model that puts the information into messages. Discrete units of information that can be passed around to the various CPU's in the architecture. Both architectures have their advantages and disadvantages.
Shared memory[edit | edit source]
Shared memory models are models where all CPU's can address the main memory of the machine. Generally the main memory is accessible though a bus. This allows all CPU's to use memory addresses, and addresses can be passed around to different CPU's. Having all CPU's access the memory through a bus introduces a number of issues. Such as bandwidth of the bus, clearly if a bus is small and there are a large number of CPU's then optimization of the CPU usage is going to be difficult to achieve. Often bus communication is augmented with local cache. Processor L1 and L2 cache, this then introduces the challenge of cache coherence. How to assure that the data in main memory and cache are up to date enough, such that all CPU's will see the appropriate data. There are several strategies to proper and optimized operation here.
Message passing systems have memory that is either not directly connected to the CPU or even possibly spread across various geographic locations. This results in a system of send and receives, where the data is packed into a message and sent across a network to various other machines in the network.
History of distributed computing[edit | edit source]
The Eighties and Nineties: PVM and MPI[edit | edit source]
Today: GFS, MapReduce, Hadoop[edit | edit source]
Hadoop[edit | edit source]
Hadoop at it's core is a combination of two open source frameworks, MapReduce and HDFS. Both are open source implementations based on white papers published by Google. MapReduce is a processing framework based on a Google framework of the same name. MapReduce takes the data stored in HDFS and processes it on each node in the cluster. MapReduce consists of two procedures defined by the programmer. The Map and the Reduce. Mappers take lines in the form of keys and values. Mappers also emit keys and values. The Reducer has an input format that is a key and a vector of all the values for that key that were emitted by the mapper. HDFS is the Hadoop filesystem. HDFS is an implementation of Google distributed filesystem. HDFS is a software implementation that distributes the storage of files across multiple nodes in a cluster. There is a default replication factor of three to assure with a high degree of certainty that no data is lost.
Mathematics of parallel processing[edit | edit source]
Parallel processing is the simultaneous execution of the same task (split up and specially adapted) on multiple processors in order to obtain faster results. The parallel nature can come from a single machine with multiple processors or multiple machines connected together to form a cluster.
Amdahl's law[edit | edit source]
Amdahl's law is a demonstration of the law of diminishing returns: while one could speed up part of a computer a hundred-fold or more, if the improvement only affects 12% of the overall task, the best the speedup could possibly be is
{\displaystyle {\frac {1}{1-0.12}}=1.136}
times faster.
More technically, the law is concerned with the speedup achievable from an improvement to a computation that affects a proportion P of that computation where the improvement has a speedup of S. (For example, if an improvement can speedup 30% of the computation, P will be 0.3; if the improvement makes the portion affected twice as fast, S will be 2.) Amdahl's law states that the overall speedup of applying the improvement will be
{\displaystyle {\frac {1}{(1-P)+{\frac {P}{S}}}}}
To see how this formula was derived, assume that the running time of the old computation was 1, for some unit of time. The running time of the new computation will be the length of time the unimproved fraction takes (which is 1 − P) plus the length of time the improved fraction takes. The length of time for the improved part of the computation is the length of the improved part's former running time divided by the speedup, making the length of time of the improved part P/S. The final speedup is computed by dividing the old running time by the new running time, which is what the above formula does.
In the special case of parallelization, Amdahl's law states that if F is the fraction of a calculation that is sequential (i.e. cannot benefit from parallelisation), and (1 − F) is the fraction that can be parallelised, then the maximum speedup that can be achieved by using N processors is
{\displaystyle {\frac {1}{F+(1-F)/N}}}
In the limit, as N tends to infinity, the maximum speedup tends to 1/F. In practice, price/performance ratio falls rapidly as N is increased once (1 − F)/N is small compared to F.
As an example, if F is only 10%, the problem can be sped up by only a maximum of a factor of 10, no matter how large the value of N used. For this reason, parallel computing is only useful for either small numbers of processors, or problems with very low values of F: so-called embarrassingly parallel problems. A great part of the craft of parallel programming consists of attempting to reduce F to the smallest possible value.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Parallel_Computing_and_Computer_Clusters/Theory&oldid=3452002"
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Introduction to Chemical Engineering Processes/Problem Solving with Multiple Components - Wikibooks, open books for an open world
1 General Strategies for Multiple-Component Operations
2 Multiple Components in a Single Operation: Separation of Ethanol and Water
General Strategies for Multiple-Component OperationsEdit
The most important thing to remember about doing mass balances with multiple components is that for each component, you can write one independent mass balance. What do I mean by independent? Well, remember we can write the general, overall mass balance for any steady-state system:
{\displaystyle \Sigma {\dot {m}}_{in}-\Sigma {\dot {m}}_{out}=0}
And we can write a similar mass balance for any component of a stream:
{\displaystyle \Sigma {\dot {m}}_{a,in}-\Sigma {\dot {m}}_{a,out}+m_{a,gen}=0}
This looks like we have three equations here, but in reality only two of them are independent because:
The sum of the masses of the components equals the total mass
The total mass generation due to reaction is always zero (by the law of mass conservation)
Therefore, if we add up all of the mass balances for the components we obtain the overall mass balance. Therefore, we can choose any set of n equations we want, where n is the number of components, but if we choose the overall mass balance as one of them we cannot use the mass balance on one of the components.
The choice of which balances to use depends on two particular criteria:
Which component(s) you have the most information on; if you don't have enough information you won't be able to solve the equations you write.
Which component(s) you can make the most reasonable assumptions about. For example, if you have a process involving oxygen and water at low temperatures and pressures, you may say that there is no oxygen dissolved in a liquid flow stream, so it all leaves by another path. This will simplify the algebra a good deal if you write the mass balance on that component.
Multiple Components in a Single Operation: Separation of Ethanol and WaterEdit
Suppose a stream containing ethanol and water (two fully miscible compounds) flows into a distillation column at 100 kg/s. Two streams leave the column: the vapor stream contains 80% ethanol by mass and the liquid bottoms has an ethanol concentration of 4M. The total liquid stream flowrate is 20 kg/s. Calculate the composition of the entrance stream.
Following the step-by-step method makes things easier.
The first step as always is to draw the flowchart, as described previously. If you do that for this system, you may end up with something like this, where x signifies mass fraction, [A] signifies molarity of A, and numbers signify stream numbers.
Now, we need to turn to converting the concentrations into appropriate units. Since the total flowrates are given in terms of mass, a unit that expresses the concentration in terms of mass of the components would be most useful. The vapor stream compositions are given as mass percents, which works well with the units of flow. However, the liquid phase concentration given in terms of a molarity is not useful for finding a mass flow rate of ethanol (or of water). Hence we must convert the concentration to something more useful.
Converting between Concentration Measurements
The easiest way to convert between concentrations is to take a careful look at the units of both what you want and what you have, and ask what physical properties (i.e. molar mass, density) you could use to interchange them. In this example, we want to convert a molarity into a mass fraction. We have from the definitions that:
{\displaystyle [A]={\frac {mol_{A}}{L_{sln}}}}
{\displaystyle x_{A}={\frac {m_{A}}{m_{sln}}}}
To convert the numerators, we need to convert moles of A to mass of A, so we can use the molar mass for this purpose. Similarly, to convert the denominators we need to change Liters to Mass, which means we'll use a density. Hence, the conversion from molarity to mass fraction is:
{\displaystyle x_{A}=[A]*{\frac {(MW)_{A}}{{\rho }_{SLN}}}}
Since we have ways to estimate
{\displaystyle {\rho }_{SLN}}
(remember them?), we can inter-convert the conversions.
In order to convert the molarity into a mass fraction, then, we need the molecular weight of ethanol and the density of a 4M ethanol solution. The former is easy if you know the chemical formula of ethanol:
{\displaystyle CH_{3}CH_{2}OH}
. Calculating the molecular weight (as you did in chem class) you should come up with about
{\displaystyle 46{\frac {g}{mol}}}
Calculating the density involves plugging in mass fractions in and of itself, so you'll end up with an implicit equation. Recall that one method of estimating a solution density is to assume that the solution is ideal (which it probably is not in this case, but if no data are available or we just want an estimate, assumptions like these are all we have, as long as we realize the values will not be exact):
{\displaystyle {\frac {1}{{\rho }_{SLN}}}=\Sigma ({\frac {x_{k}}{{\rho }_{k}}})}
In this case, then,
{\displaystyle {\frac {1}{{\rho }_{SLN}}}={\frac {x_{EtOH}}{{\rho }_{EtOH}}}+{\frac {x_{H2O}}{{\rho }_{H2O}}}}
We can look up the densities of pure water and pure ethanol, they are as follows (from Wikipedia's articles w:Ethanol and w:Water):
{\displaystyle {\rho }_{EtOH}=0.789{\frac {g}{cm^{3}}}=789{\frac {g}{L}}}
{\displaystyle {\rho }_{H2O}=1.00{\frac {g}{cm^{3}}}=1000{\frac {g}{L}}}
Therefore, since the mass fractions add to one, our equation for density becomes:
{\displaystyle {\frac {1}{{\rho }_{sln}}}={\frac {x_{EtOH}}{789{\frac {g}{L}}}}+{\frac {1-x_{EtOH}}{1000{\frac {g}{L}}}}}
From the NOTE above, we can now finally convert the molarity into a mass fraction as:
{\displaystyle x_{EtOH}=[EtOH]*{\frac {(MW)_{EtOH}}{{\rho }_{SLN}}}=4{\frac {mol}{L}}*46{\frac {g}{mol}}*({\frac {x_{EtOH}}{789{\frac {g}{L}}}}+{\frac {1-x_{EtOH}}{1000{\frac {g}{L}}}})}
{\displaystyle x_{EtOH}=0.194}
Since we are seeking properties related to mass flow rates, we will need to relate our variables with mass balances.
Remember that we can do a mass balance on any of the N independent species and one on the overall mass, but since the sum of the individual masses equals the overall only
{\displaystyle N-1}
of these equations will be independent. It is often easiest mathematically to choose the overall mass balance and
{\displaystyle N-1}
individual species balances, since you don't need to deal with concentrations for the overall measurements.
Since our concentrations are now in appropriate units, we can do any two mass balances we want. Lets choose the overall first:
{\displaystyle {\dot {m}}_{1}-{\dot {m}}_{2}-{\dot {m}}_{3}=0}
Plugging in known values:
{\displaystyle {\dot {m}}_{2}=100{\frac {kg}{s}}-20{\frac {kg}{s}}}
{\displaystyle {\dot {m}}_{2}=80{\frac {kg}{s}}}
{\displaystyle {\dot {m}}_{2}}
we can do a mass balance on either ethanol or water to find the composition of the input stream. Lets choose ethanol (A):
{\displaystyle {\dot {m}}_{A1}={\dot {m}}_{A2}+{\dot {m}}_{A3}}
Written in terms of mass fractions this becomes:
{\displaystyle x_{A1}*{\dot {m}}_{1}=x_{A2}*{\dot {m}}_{2}+x_{A3}*{\dot {m}}_{3}}
{\displaystyle x_{A1}*100{\frac {kg}{s}}=0.8*80{\frac {kg}{s}}+0.194*20{\frac {kg}{s}}}
{\displaystyle x_{A1}=0.68}
Hence, the feed is 68% Ethanol and 32% Water.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Introduction_to_Chemical_Engineering_Processes/Problem_Solving_with_Multiple_Components&oldid=3325792"
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Introduction to Chemical Engineering Processes/Reactions with Recycle - Wikibooks, open books for an open world
1 Introduction to Reactions with Recycle
2 Example Reactor with Recycle
2.1 DOF Analysis
2.2 Plan and Solution
2.3 Reactor Analysis
2.4 Comparison to the situation without the separator/recycle system
Introduction to Reactions with RecycleEdit
Reactions with recycle are very useful for a number of reasons, most notably because they can be used to improve the selectivity of multiple reactions, push a reaction beyond its equilibrium conversion, or speed up a catalytic reaction by removing products. A recycle loop coupled with a reactor will generally contain a separation process in which unused reactants are (partially) separated from products. These reactants are then fed back into the reactor along with the fresh feed.
Example Reactor with RecycleEdit
Consider a system designed for the hydrogenation of ethylene into ethane:
{\displaystyle 2H_{2}+C_{2}H_{2}\rightarrow C_{2}H_{6}}
{\displaystyle (2A+B\rightarrow C)}
The reaction takes too long to go to completion (and releases too much heat) so the designers decided to implement a recycle system in which, after only part of the reaction had finished, the mixture was sent into a membrane separator. There, most of the ethylene was separated out, with little hydrogen or ethylene contamination. After this separation, the cleaned stream entered a splitter, where some of the remaining mixture was returned to the reactor and the remainder discarded.
The system specifications for this process were as follows:
Feed: 584 kg/h ethylene, 200 kg/h hydrogen gas
Outlet stream from reactor contains 15% hydrogen by mass
Mass flows from membrane separator: 100 kg/h, 5% Hydrogen and 93% ethane
Splitter: 30% reject and 70% reflux
What was the extent of reaction for this system? What would the extent of reaction be if there was no separation/recycle process after (assume that the mass percent of hydrogen leaving the reactor is the same)? What limits how effective this process can be?
Let's first draw our flowchart as usual:
DOF AnalysisEdit
On reactor: 6 unknowns
{\displaystyle ({\dot {m}}_{5},x_{A5},x_{B5},{\dot {m}}_{3},x_{B3},X)}
- 3 equations = 3 DOF
On separator: 5 unknowns
{\displaystyle ({\dot {m}}_{3},x_{B3},{\dot {m}}_{5},x_{A5},x_{B5})}
On splitter: 3 unknowns - 0 equations (we used all of them in labeling the chart) -> 3 DOF
Duplicate variables: 8 (
{\displaystyle {\dot {m}}_{5},x_{A5},x_{B5}}
twice each and
{\displaystyle {\dot {m}}_{3},x_{B3}}
Total DOF = 8 - 8 = 0 DOF
Plan and SolutionEdit
Generally, though not always, it is easiest to deal with the reactor itself last because it usually has the most unknowns. Lets begin by looking at the overall system because we can often get some valuable information from that.
Overall System DOF(overall system) = 4 unknowns (
{\displaystyle {\dot {m}}_{5},x_{A5},X,x_{B5}}
) - 3 equations = 1 DOF.
We CANNOT say that total mass of A and B is conserved because we have a reaction here! Therefore we must include the conversion X in our list of unknowns for both the reactor and the overall system. However, the total mass in the system is conserved so we can solve for
{\displaystyle {\dot {m}}_{5}}
Let's go ahead and solve for m5 though because that'll be useful later.
{\displaystyle 784=100+0.3({\dot {m}}_{5})}
{\displaystyle {\dot {m}}_{5}=2280{\mbox{ kg/h}}}
We can't do anything else with the overall system without knowing the conversion so lets look elsewhere.
DOF(separator) = 4 unknowns (
{\displaystyle {\dot {m}}_{3},x_{B3},x_{A5},x_{B5}}
) - 3 equations = 1 DOF. Let's solve for those variables we can though.
We can solve for m3 because from the overall material balance on the separator:
{\displaystyle {\dot {m}}_{3}={\dot {m}}_{4}+{\dot {m}}_{5}}
{\displaystyle {\dot {m}}_{3}=100+2280}
{\displaystyle {\dot {m}}_{3}=2380{\mbox{ kg/h}}}
Then we can do a mass balance on A to solve for xA5:
{\displaystyle {\dot {m}}_{3}x_{A3}={\dot {m}}_{4}x_{A4}+{\dot {m}}_{5}x_{A5}}
{\displaystyle 2380(0.15)=100(0.05)+2380(x_{A5})}
{\displaystyle x_{A5}=.1544}
{\displaystyle x_{B5}}
{\displaystyle x_{B3}}
, we cannot use the mass balance on B or C for the separator, so lets move on. Let's now turn to the reactor:
Reactor AnalysisEdit
DOF: 3 unknowns remaining (
{\displaystyle x_{B3},x_{B5},andX}
) - 2 equations (because the overall balance is already solved!) = 1 DOF. Therefore we still cannot solve the reactor completely. However, we can solve for the conversion and generation terms given what we know at this point. Lets start by writing a mole balance on A in the reactor.
{\displaystyle {\dot {n}}_{A1}+{\dot {n}}_{A,recycle}-X*a={\dot {n}}_{A3}}
To find the three nA terms we need to convert from mass to moles (since A is hydrogen, H2, the molecular weight is
{\displaystyle {\frac {1{\mbox{ mol}}}{0.002016{\mbox{ g}}}}}
{\displaystyle {\dot {n}}_{A1}=200{\frac {kg}{h}}*{\frac {1{\mbox{ mol}}}{0.002016{\mbox{ kg}}}}=99206{\frac {\mbox{mol A}}{h}}}
{\displaystyle {\dot {n}}_{A,recycle}=0.7*{\frac {m_{5}*x_{A5}}{MW_{A}}}={\frac {0.7(2280)(0.1544)}{0.002016}}=122000{\frac {\mbox{mol A}}{h}}}
Thus the total amount of A entering the reactor is:
{\displaystyle {\dot {n}}_{A,in}=99206+122000=221428{\frac {\mbox{mol A}}{h}}}
The amount exiting is:
{\displaystyle {\dot {n}}_{A,out}={\frac {{\dot {m}}_{3}*x_{A3}}{MM_{A}}}={\frac {2380*0.15}{0.002016}}=177083{\frac {\mbox{mol A}}{h}}}
Therefore we have the following from the mole balance:
{\displaystyle 221428-2X=177083}
{\displaystyle X=22173{\frac {\mbox{ moles}}{h}}}
Now that we have this we can calculate the mass of B and C generated:
{\displaystyle m_{B,gen}=-Xb*MW_{B}=22173{\frac {\mbox{ mol B}}{h}}*0.026{\frac {kg}{\mbox{ mol B}}}=-576.5{\frac {\mbox{ kg B}}{h}}}
{\displaystyle m_{C,gen}=+Xc*MW_{C}=22173{\frac {\mbox{ mol C}}{h}}*0.030{\frac {kg}{\mbox{ mol C}}}=+665.2{\frac {\mbox{ kg C}}{h}}}
At this point you may want to calculate the amount of B and C leaving the reactor with the mass balances on B and C:
{\displaystyle 584+0.7*x_{B5}*2280-576.5=x_{B3}*2380}
{\displaystyle 0.7*(1-0.1544-x_{B5})*2280+665.2=(1-0.15-x_{B3})*2380}
However, these equations are exactly the same! Therefore, we have proven our assertion that there is still 1 DOF in the reactor. So we need to look elsewhere for something to calculate xB5. That place is the separator balance on B:
{\displaystyle {\dot {m}}_{3}*x_{B3}={\dot {m}}_{4}*x_{B4}+{\dot {m}}_{5}*x_{B5}}
{\displaystyle 2380x_{B3}=0.02(100)+2280x_{B5}}
Solving these two equations (1) and (2) yields the final two variables in the system:
{\displaystyle x_{B3}=0.00856,x_{B5}=0.008058}
Note that this means the predominant species in stream 5 is also C (
{\displaystyle x_{C5}=0.838}
). However, the separator/recycle setup does make a big difference, as we'll see next.
Comparison to the situation without the separator/recycle systemEdit
Now that we know how much ethane we can obtain from the reactor after separating, let's compare to what would happen without any of the recycle systems in place. With the same data as in the first part of this problem, the new flowchart looks like this:
There are three unknowns (
{\displaystyle {\dot {m}}_{3},x_{B3},X}
) and three independent material balances, so the problem can be solved. Starting with an overall mass balance because total mass is conserved:
{\displaystyle {\dot {m}}_{1}+{\dot {m}}_{2}={\dot {m}}_{3}}
{\displaystyle {\dot {m}}_{3}=789{\frac {kg}{h}}}
We can carry out the same sort of analysis on the reactor as we did in the previous section to find the conversion and mass percent of C in the exit stream, which is left as an exercise to the reader. The result is that:
{\displaystyle X=20250{\mbox{ moles}},x_{C3}=0.77}
Compare this to the two exit streams in the recycle setup. Both of the streams were richer in C than 77%, even the reject stream. This occurred because the unreacted A and B was allowed to re-enter the reactor and form more C, and the separator was able to separate almost all the C that formed from the unreacted A and B.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Introduction_to_Chemical_Engineering_Processes/Reactions_with_Recycle&oldid=3502692"
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https://stats.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fstats.libretexts.org%2FBookshelves%2FIntroductory_Statistics%2FBook%253A_Statistical_Thinking_for_the_21st_Century_(Poldrack)%2F10%253A_Probability%2F10.03%253A_Appendix
Proof (Derivation of Bayes’ rule). First, remember the rule for computing a conditional probability:
P(A|B) = \frac{P(A \cap B)}{P(B)}
P(A \cap B) = P(A|B) * P(B)
P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A|B)*P(B)}{P(A)}
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convert 1 joule to ergs in dimensional analysis - Physics - Units And Measurements - 9104569 | Meritnation.com
\mathrm{Joule} \mathrm{is} \mathrm{the} \mathrm{unit} \mathrm{for} \mathrm{the} \mathrm{quantity} \mathrm{of} \mathrm{energy}.\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\phantom{\rule{0ex}{0ex}}\mathrm{Dimensional} \mathrm{formula} \mathrm{for} \mathrm{joule} \mathrm{is} \left[{\mathrm{M}}^{1}{\mathrm{L}}^{2}{\mathrm{T}}^{-2}\right]\phantom{\rule{0ex}{0ex}}\mathrm{Let}\phantom{\rule{0ex}{0ex}}{\mathrm{n}}_{1}\left(\mathrm{joule}\right)={\mathrm{n}}_{2}\left(\mathrm{erg}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\phantom{\rule{0ex}{0ex}}{\mathrm{n}}_{1}\left[{{\mathrm{M}}_{1}}^{1}{{\mathrm{L}}_{1}}^{2}{{\mathrm{T}}_{1}}^{-2}\right]={\mathrm{n}}_{2}\left[{{\mathrm{M}}_{2}}^{1}{{\mathrm{L}}_{2}}^{2}{{\mathrm{T}}_{2}}^{-2}\right] \mathrm{Here} 1-> \mathrm{SI} \mathrm{system} \mathrm{and} 2->\mathrm{CGS} \mathrm{system}\phantom{\rule{0ex}{0ex}}{\mathrm{n}}_{2}={\mathrm{n}}_{1}\frac{\left[{{\mathrm{M}}_{1}}^{1}{{\mathrm{L}}_{1}}^{2}{{\mathrm{T}}_{1}}^{-2}\right]}{\left[{{\mathrm{M}}_{2}}^{1}{{\mathrm{L}}_{2}}^{2}{{\mathrm{T}}_{2}}^{-2}\right]}\phantom{\rule{0ex}{0ex}} ={\mathrm{n}}_{1}×\left(\frac{{\mathrm{M}}_{1}}{{\mathrm{M}}_{2}}\right)×{\left(\frac{{\mathrm{L}}_{1}}{{\mathrm{L}}_{2}}\right)}^{2}×{\left(\frac{{\mathrm{T}}_{1}}{{\mathrm{T}}_{2}}\right)}^{-2}\phantom{\rule{0ex}{0ex}} =1 ×\left(\frac{1 \mathrm{kg}}{1 \mathrm{g}}\right)×{\left(\frac{1 \mathrm{m}}{1 \mathrm{cm}}\right)}^{2}×{\left(\frac{1 \mathrm{s}}{1 \mathrm{s}}\right)}^{-2}\phantom{\rule{0ex}{0ex}} =1 ×\left(\frac{1000 \mathrm{g}}{1 \mathrm{g}}\right)×{\left(\frac{100 \mathrm{cm}}{1 \mathrm{cm}}\right)}^{2}×{\left(\frac{1 \mathrm{s}}{1 \mathrm{s}}\right)}^{-2}\phantom{\rule{0ex}{0ex}} ={10}^{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\phantom{\rule{0ex}{0ex}}{\mathrm{n}}_{1}\left(\mathrm{joule}\right)={\mathrm{n}}_{2}\left(\mathrm{erg}\right)\phantom{\rule{0ex}{0ex}}\mathbf{1}\mathbf{ }\mathbf{joule}\mathbf{ }\mathbf{=}\mathbf{ }{\mathbf{10}}^{\mathbf{7}}\mathbf{ }\mathbf{ergs}\phantom{\rule{0ex}{0ex}}
Yuvraj Saini answered this
10000000 erg = i joule
Yamini Bisht answered this
thnkss..,.but I want dimensional analysis
Aman Bhatt answered this
Joule dimension is same of momentom
momentom = mv ( where m is mass and v is velocity )
= M*LT =MLT
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Introduction to Chemical Engineering Processes/Why use mole balances? - Wikibooks, open books for an open world
1 Review of Reaction Stoichiometry
2 Molecular Mole Balances
3 Extent of Reaction
4 Mole Balances and Extents of Reaction
Review of Reaction StoichiometryEdit
Up until now, all of the balances we have done on systems have been in terms of mass. However, mass is inconvenient for a reacting system because it does not allow us to take advantage of the stoichiometry of the reaction in relating the relative amounts of reactants and of products.
Stoichiometry is the relationship between reactants and products in a balanced reaction as given by the ratio of their coefficients. For example, in the reaction:
{\displaystyle C_{2}H_{2}+2H_{2}\rightarrow C_{2}H_{6}}
the reaction stoichiometry would dictate that for every one molecule of
{\displaystyle C_{2}H_{2}}
(acetylene) that reacts, two molecules of
{\displaystyle H_{2}}
(hydrogen) are consumed and one molecule of
{\displaystyle C_{2}H_{6}}
are formed. However, this does not hold for grams of products and reactants.
Even though the number of molecules in single substance is proportional to the mass of that substance, the constant of proportionality (the molecular mass) is not the same for every molecule. Hence, it is necessary to use the molecular weight of each molecule to convert from grams to moles in order to use the reaction's coefficients.
Molecular Mole BalancesEdit
We can write balances on moles like we can on anything else. We'll start with our ubiquitous general balance equation:
{\displaystyle Input-Output=Accumulation-Generation}
As usual we assume that accumulation = 0 in this book so that:
{\displaystyle Input-Output+Generation=0}
Let us denote molar flow rates by
{\displaystyle {\dot {n}}}
to distinguish them from mass flow rates. We then have a similar equation to the mass balance equation:
{\displaystyle \Sigma {\dot {n}}_{in}-\Sigma {\dot {n}}_{out}+n_{gen}=0}
The same equation can be written in terms of each individual species.
There are a couple of important things to note about this type of balance as opposed to a mass balance:
Just like with the mass balance, in a mole balance, a non-reactive system has
{\displaystyle n_{gen}=0}
Unlike the mass balance, the TOTAL generation of moles isn't necessarily 0 even for the overall mole balance! To see this, consider how the total number of moles changes in the above reaction; the final number of moles will not equal the initial number because 3 total moles of molecules are reacting to form 1 mole of products.
Why would we use it if the generation isn't necessarily 0? We use the molecular mole balance because if we know how much of any one substance is consumed or created in the reaction, we can find all of the others from the reaction stoichiometry. This is a very powerful tool because each reaction only creates one new unknown if you use this method! The following section is merely a formalization of this concept, which can be used to solve problems involving reactors.
Extent of ReactionEdit
In order to formalize the previous analysis of reactions in terms of a single variable, let us consider the generic reaction:
{\displaystyle aA+bB\rightarrow cC+dD}
The Molar Extent of Reaction X is defined as:
{\displaystyle X=-{\frac {\Delta n_{A}}{a}}=-{\frac {\Delta n_{B}}{b}}={\frac {\Delta n_{C}}{c}}={\frac {\Delta n_{D}}{d}}}
Since all of these are equivalent, it is possible to find the change in moles of any species in a given reaction if the extent of reaction X is known.
Though they won't be discussed here, there are other ways in which the extent of reaction can be defined. Some other definitions are dependent on the percent change of a particular substrate, and the stoichiometry is used in a different way to determine the change in the others. This definition makes X independent of the substrate you choose.
The following example illustrates the use of the extent of reaction.
Consider the reaction
{\displaystyle H_{2}O_{2}+O_{3}\rightarrow H_{2}O+2O_{2}}
. If you start with 50 g of
{\displaystyle H_{2}O_{2}}
and 25 grams of
{\displaystyle O_{3}}
, and 25% of the moles of
{\displaystyle O_{3}}
are consumed, find the molar extent of reaction and the changes in the other components.
Solution: First we need to convert to moles, since stoichiometry is not valid when units are in terms of mass.
{\displaystyle 50{\mbox{ g H}}_{2}O_{2}*{\frac {1{\mbox{ mol}}}{34{\mbox{ g}}}}=1.471{\mbox{ moles H}}_{2}O_{2}}
{\displaystyle 25{\mbox{ g O}}_{3}*{\frac {1{\mbox{ mol}}}{48{\mbox{ g}}}}=0.5208{\mbox{ moles O}}_{3}}
Clearly ozone is the limiting reactant here. Since 25% is consumed, we have that:
{\displaystyle \Delta n(O_{3})=-0.25*0.5208=-0.1302{\mbox{ moles O}}_{3}}
Hence, by definition,
{\displaystyle X={\frac {-0.1302}{1}}=0.1302}
{\displaystyle \Delta n(H_{2}O_{2})=-0.1302,\Delta n(H_{2}O)=0.1302,\Delta n(O_{2})=2*0.1302=0.2604}
, all in moles of the appropriate substrate.
Mole Balances and Extents of ReactionEdit
The mole balance written above can be written in terms of extent of reaction if we notice that the
{\displaystyle \Delta n(A)}
term defined above is exactly the number of moles of a generated or consumed by the reaction.
This is only useful for individual species balances, not the overall mole balance. When doing balances on reactive systems, unlike with non-reactive systems, it is generally easier to use all individual species balances possible, rather than the total mole balance and then all but one of the individual species. This is because the total generation of moles in a reaction is generally not 0, so no algebraic advantage is gained by using the total material balance on the system.
Therefore we can write that:
{\displaystyle n_{A,gen}=\Delta n(A)=-X*a}
where X is the molar extent of reaction and a is the stoichiometric coefficient of A. Plugging this into the mole balance derived earlier, we arrive at the molecular mole balance equation:
{\displaystyle \Sigma {\dot {n}}_{A,out}-\Sigma {n}_{A,in}-X*a=0}
if A is consumed, or +Xa if it is generated in the reaction
Retrieved from "https://en.wikibooks.org/w/index.php?title=Introduction_to_Chemical_Engineering_Processes/Why_use_mole_balances%3F&oldid=3325808"
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Base and Normality - Maple Help
Home : Support : Online Help : Math Apps : Real and Complex Numbers : Base and Normality
Base and Normality
A numeral system is a way of representing real numbers as an ordered sequence of symbols called digits from a finite ordered set. The number,
b
, of symbols in this set is called the base. The symbols themselves represent the number zero, followed by the first
b-1
positive integers.
For any base
b
, any real number can be written as a sum of the form
\underset{i=-\mathrm{∞}}{\overset{n}{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∑}}}\phantom{\rule[-0.0ex]{5.0px}{0.0ex}}{x}_{i}\mathbf{}{b}^{i}
0 ≤ {x}_{i}<b
The corresponding base
b
representation of this number is:
{x}_{n}\mathbf{}{x}_{n-1}\cdot \cdot \cdot {x}_{1}\mathbf{}{x}_{0\mathbf{·}}{x}_{-1}{x}_{-2}\mathbf{}\cdot \cdot \cdot
The standard numeral system around the world is the base ten decimal system, which uses the digits {0,1,2,3,4,5,6,7,8,9}. Systems using a base other than ten are used commonly in computing, including:
binary (base two), binary digits (bits) = {0,1}
octal (base eight), octal digits = {0,1,2,3,4,5,6,7}
hexadecimal (base sixteen), hexadecimal digits = {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F}
In number theory, a real number
\mathbit{x}
is called normal in base
\mathbit{b}
if the sequence of digits in its representation in base
b
appears random, in the following sense: The density of any length
k
digit subsequence
{x}_{i+1}\cdot \cdot \cdot {x}_{i+k}
in the representation of
x
{b}^{-k}
x
is normal if it is normal in every base
b>1
Input a Maple expression in the box below (or choose one from the drop-down box) that evaluates to a real number. Choose a base b > 1, and Maple will find the base b representation for your number. Use the slider to adjust the number of significant figures, and look at the graph to see if your number is normal in that base.
Pi22/7sqrt(2)exp(1)ln(3)Other
# of significant figures =
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London dispersion force - Wikipedia
Cohesive force between species
Find sources: "London dispersion force" – news · newspapers · books · scholar · JSTOR (August 2018) (Learn how and when to remove this template message)
2 Quantum mechanical theory
3 Relative magnitude
While the London dispersion force between individual atoms and molecules is quite weak and decreases quickly with separation (R) like
{\displaystyle {\frac {1}{R^{6}}}}
, in condensed matter (liquids and solids), the effect is cumulative over the volume of materials,[5] or within and between organic molecules, such that London dispersion forces can be quite strong in bulk solid and liquids and decay much more slowly with distance. For example, the total force per unit area between two bulk solids decreases by
{\displaystyle {\frac {1}{R^{3}}}}
[6] where R is the separation between them. The effects of London dispersion forces are most obvious in systems that are very non-polar (e.g., that lack ionic bonds), such as hydrocarbons and highly symmetric molecules like bromine (Br2, a liquid at room temperature) or iodine (I2, a solid at room temperature). In hydrocarbons and waxes, the dispersion forces are sufficient to cause condensation from the gas phase into the liquid or solid phase. Sublimation heats of e.g. hydrocarbon crystals reflect the dispersion interaction. Liquification of oxygen and nitrogen gases into liquid phases is also dominated by attractive London dispersion forces.
Quantum mechanical theory[edit]
London wrote a Taylor series expansion of the perturbation in
{\displaystyle {\frac {1}{R}}}
{\displaystyle R}
is the distance between the nuclear centers of mass of the moieties.
In this manner, the following approximation is obtained for the dispersion interaction
{\displaystyle E_{AB}^{\rm {disp}}}
between two atoms
{\displaystyle A}
{\displaystyle B}
{\displaystyle \alpha _{A}}
{\displaystyle \alpha _{B}}
are the dipole polarizabilities of the respective atoms. The quantities
{\displaystyle I_{A}}
{\displaystyle I_{B}}
are the first ionization potentials of the atoms, and
{\displaystyle R}
is the intermolecular distance.
{\displaystyle E_{AB}^{\rm {disp}}\approx -{3 \over 2}{I_{A}I_{B} \over I_{A}+I_{B}}{\alpha _{A}\alpha _{B} \over {R^{6}}}}
Relative magnitude[edit]
Retrieved from "https://en.wikipedia.org/w/index.php?title=London_dispersion_force&oldid=1087970864"
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Assemble finite element matrices - MATLAB assembleFEMatrices - MathWorks Deutschland
Finite Element Matrices for 2-D Problem
Specified Set of Finite Element Matrices
Finite Element Matrices with nullspace and stiff-spring Methods
Finite Element Matrices for Time-Dependent Problem
Finite Element Matrices for Nonlinear Problem
bcmethod
Imposing Dirichlet Boundary Conditions
Thermal, Structural, and Electromagnetic Analysis
Assemble finite element matrices
FEM = assembleFEMatrices(model)
FEM = assembleFEMatrices(model,matrices)
FEM = assembleFEMatrices(model,bcmethod)
FEM = assembleFEMatrices(___,state)
FEM = assembleFEMatrices(model) returns a structural array containing all finite element matrices for a PDE problem specified as a model.
FEM = assembleFEMatrices(model,matrices) returns a structural array containing only the specified finite element matrices.
FEM = assembleFEMatrices(model,bcmethod) assembles finite element matrices and imposes boundary conditions using the method specified by bcmethod.
FEM = assembleFEMatrices(___,state) assembles finite element matrices using the input time or solution specified in the state structure array. The function uses the time field of the structure for time-dependent models and the solution field u for nonlinear models. You can use this argument with any of the previous syntaxes.
Create a PDE model for the Poisson equation on an L-shaped membrane with zero Dirichlet boundary conditions.
applyBoundaryCondition(model,'Edge',1:model.Geometry.NumEdges, ...
Generate a mesh and obtain the default finite element matrices for the problem and mesh.
FEM = struct with fields:
A: [401x401 double]
F: [401x1 double]
Q: [401x401 double]
G: [401x1 double]
H: [80x401 double]
Make computations faster by specifying which finite element matrices to assemble.
Create a transient thermal model and include the geometry of the built-in function squareg.
Specify the thermal conductivity of the material and the internal heat source.
thermalProperties(thermalmodel,'ThermalConductivity',0.2);
thermalBC(thermalmodel,'Edge',[1,3],'Temperature',100);
Assemble the stiffness and mass matrices.
FEM_KM = assembleFEMatrices(thermalmodel,'KM')
FEM_KM = struct with fields:
Now, assemble the finite element matrices M, K, A, and F.
FEM_MKAF = assembleFEMatrices(thermalmodel,'MKAF')
FEM_MKAF = struct with fields:
The four matrices M, K, A, and F correspond to discretized versions of the PDE coefficients m, c, a, and f. These four matrices also represent the domain of the finite-element model of the PDE. Instead of specifying them explicitly, you can use the domain argument.
FEMd = assembleFEMatrices(thermalmodel,'domain')
FEMd = struct with fields:
The four matrices Q, G, H, and R, correspond to discretized versions of q, g, h, and r in the Neumann and Dirichlet boundary condition specification. These four matrices also represent the boundary of the finite-element model of the PDE. Use the boundary argument to assemble only these matrices.
FEMb = assembleFEMatrices(thermalmodel,'boundary')
FEMb = struct with fields:
H: [74x1541 double]
G: [1541x1 double]
Q: [1541x1541 double]
Obtain the finite element matrices after imposing the boundary condition using the null-space approach. This approach eliminates the Dirichlet degrees of freedom and provides a reduced system of equations.
FEMn = assembleFEMatrices(model,'nullspace')
FEMn = struct with fields:
Kc: [321x321 double]
Fc: [321x1 double]
B: [401x321 double]
ud: [401x1 double]
Obtain the solution to the PDE using the nullspace finite element matrices.
un = FEMn.B*(FEMn.Kc\FEMn.Fc) + FEMn.ud;
Compare this result to the solution given by solvepde. The two solutions are identical.
u1 = solvepde(model);
norm(un - u1.NodalSolution)
Obtain the finite element matrices after imposing the boundary condition using the stiff-spring approach. This approach retains the Dirichlet degrees of freedom, but imposes a large penalty on them.
FEMs = assembleFEMatrices(model,'stiff-spring')
FEMs = struct with fields:
Ks: [401x401 double]
Fs: [401x1 double]
Obtain the solution to the PDE using the stiff-spring finite element matrices. This technique gives a less accurate solution.
us = FEMs.Ks\FEMs.Fs;
norm(us - u1.NodalSolution)
Assemble finite element matrices for the first and last time steps of a transient structural problem.
Create a transient structural model for solving a solid (3-D) problem.
gm = multicylinder(0.01,0.05);
addVertex(gm,'Coordinates',[0,0,0.05]);
Specify that the bottom of the cylinder is a fixed boundary.
Specify the harmonic pressure on the top of the cylinder.
structuralBoundaryLoad(structuralmodel,'Face',2,...
Generate a linear mesh.
generateMesh(structuralmodel,'GeometricOrder','linear');
Assemble the finite element matrices for the initial time step.
FEM_domain = assembleFEMatrices(structuralmodel,state)
FEM_domain = struct with fields:
H: [252x6609 double]
Pressure applied at the top of the cylinder is the only time-dependent quantity in the model. To model the dynamics of the system, assemble the boundary-load finite element matrix G for the initial, intermediate, and final time steps.
FEM_boundary_init = assembleFEMatrices(structuralmodel,'G',state)
FEM_boundary_init = struct with fields:
state.time = tlist(floor(length(tlist)/2));
FEM_boundary_half = assembleFEMatrices(structuralmodel,'G',state)
FEM_boundary_half = struct with fields:
state.time = tlist(end);
FEM_boundary_final = assembleFEMatrices(structuralmodel,'G',state)
FEM_boundary_final = struct with fields:
Assemble finite element matrices for a heat transfer problem with temperature-dependent thermal conductivity.
Convert the decsg format into a geometry object. Include the geometry in the model and plot the geometry.
Set the temperature on the left edge to 100 degrees. Set the heat flux out of the block on the right edge to -10. The top and bottom edges and the edges inside the cavity are all insulated: there is no heat transfer across these edges.
Specify the thermal conductivity of the material as a simple linear function of temperature u.
generateMesh(thermalmodelS);
Rnonlin = solve(thermalmodelS);
Because the thermal conductivity is nonlinear (depends on the temperature), compute the system matrices corresponding to the converged temperature. Assign the temperature distribution to the u field of the state structure array. Because the u field must contain a row vector, transpose the temperature distribution.
state.u = Rnonlin.Temperature.';
Assemble finite element matrices using the temperature distribution at the nodal points.
FEM = assembleFEMatrices(thermalmodelS,'nullspace',state)
Kc: [1277x1277 double]
Fc: [1277x1 double]
B: [1320x1277 double]
ud: [1320x1 double]
Compute the solution using the system matrices to verify that they yield the same temperature as Rnonlin.
u = FEM.B*(FEM.Kc\FEM.Fc) + FEM.ud;
Compare this result to the solution given by solve.
norm(u - Rnonlin.Temperature)
assembleFEMatrices does not support assembling FE matrices for 3-D magnetostatic analysis models.
Example: emagmodel = createpde('electromagnetic','electrostatic')
bcmethod — Method for including boundary conditions
'none' (default) | 'nullspace' | 'stiff-spring'
Method for including boundary conditions, specified as 'none', 'nullspace', or 'stiff-spring'. For more information, see Algorithms.
Example: FEM = assembleFEMatrices(model,'nullspace')
matrices — Matrices to assemble
matrix identifiers | 'boundary' | 'domain'
Matrices to assemble, specified as:
Matrix identifiers, such as 'F', 'MKF', 'K', and so on — Assemble the corresponding matrices. Each uppercase letter represents one matrix: K, A, F, Q, G, H, R, M, and T. You can combine several letters into one character vector or string, such as 'MKF'.
'boundary' — Assemble all matrices related to geometry boundaries.
'domain' — Assemble all domain-related matrices.
Example: FEM = assembleFEMatrices(model,'KAF')
state — Time for time-dependent models and solution for nonlinear models
Time for time-dependent models and solution for nonlinear models, specified in a structure array. The array fields represent the following values:
state.time contains a nonnegative number specifying the time value for time-dependent models.
state.u contains a solution matrix of size N-by-Np that can be used to assemble matrices in a nonlinear problem setup, where coefficients are functions of state.u. Here, N is the number of equations in the system, and Np is the number of nodes in the mesh.
Example: state.time = tlist(end); FEM = assembleFEMatrices(model,'boundary',state)
FEM — Finite element matrices
structural array
Finite element matrices, returned as a structural array. Use the bcmethod and matrices arguments to specify which finite element matrices you want to assemble.
The fields in the structural array depend on bcmethod:
If the value is 'none', then the fields are K, A, F, Q, G, H, R, and M.
If the value is 'nullspace', then the fields are Kc, Fc, B, ud, and M.
If the value is 'stiff-spring', then the fields are Ks, Fs, and M.
The fields in the structural array also depend on matrices:
If the value is boundary, then the fields are all matrices related to geometry boundaries.
If the value is domain, then the fields are all domain-related matrices.
If the value is a matrix identifier or identifiers, such as 'F', 'MKF', 'K', and so on, then the fields are the corresponding matrices.
m\frac{{\partial }^{2}u}{\partial {t}^{2}}+d\frac{\partial u}{\partial t}-\nabla ·\left(c\otimes \nabla u\right)+au=f
\begin{array}{l}-\nabla ·\left(c\otimes \nabla u\right)+au=\lambda du\\ \text{or}\\ -\nabla ·\left(c\otimes \nabla u\right)+au={\lambda }^{2}mu\end{array}
with the Dirichlet boundary conditions, hu = r, and Neumann boundary conditions,
n\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(c\otimes \nabla u\right)+qu=g
assembleFEMatrices returns the following full finite element matrices and vectors that represent the corresponding PDE problem:
K is the stiffness matrix, the integral of the discretized version of the c coefficient.
M is the mass matrix, the integral of the discretized version of the m or d coefficients. M is nonzero for time-dependent and eigenvalue problems.
A is the integral of the discretized version of the a coefficient.
F is the integral of the discretized version of the f coefficient. For thermal, electromagnetic, and structural problems, F is a source or body load vector.
Q is the integral of the discretized version of the q term in a Neumann boundary condition.
G is the integral of the discretized version of the g term in a Neumann boundary condition. For structural problems, G is a boundary load vector.
The H and R matrices come directly from the Dirichlet conditions and the mesh.
The 'nullspace' technique eliminates Dirichlet conditions from the problem using a linear algebra approach. It generates the combined finite-element matrices Kc, Fc, B, and vector ud corresponding to the reduced system Kc*u = Fc, where Kc = B'*(K + A + Q)*B, and Fc = B'*(F + G). The B matrix spans the null space of the columns of H (the Dirichlet condition matrix representing h*ud = r). The R vector represents the Dirichlet conditions in H*ud = R. The ud vector has the size of the solution vector. Its elements are zeros everywhere except at Dirichlet degrees-of-freedom (DoFs) locations where they contain the prescribed values.
From the 'nullspace' matrices, you can compute the solution u as
u = B*(Kc\Fc) + ud.
If you assembled a particular set of matrices, for example G and M, you can impose the boundary conditions on G and M as follows. First, compute the nullspace of columns of H.
[B,Or] = pdenullorth(H);
ud = Or*((H*Or\R)); % Vector with known value of the constraint DoF.
Then use the B matrix as follows. To eliminate Dirichlet degrees of freedom from the load vector G, use:
GwithBC = B'*G
To eliminate Dirichlet degrees of freedom from mass matrix, use:
M = B'*M*B
You can eliminate Dirichlet degrees of freedom from other vectors and matrices using the same technique.
The 'stiff-spring' technique converts Dirichlet boundary conditions to Neumann boundary conditions using a stiff-spring approximation. It returns a matrix Ks and a vector Fs that together represent a different type of combined finite element matrices. The approximate solution is u = Ks\Fs. Compared to the 'nullspace' technique, the 'stiff-spring' technique generates matrices more quickly, but generally gives less accurate solutions.
Internally, the toolbox uses the 'nullspace' approach to impose Dirichlet boundary conditions while computing the solution using solvepde and solve.
If the number of nodes in a model is NumNodes, and the number of equations is N, then the length of column vectors u and ud is N*NumNodes. The toolbox assigns the IDs to the degrees of freedom in u and ud:
Entries from 1 to NumNodes correspond to the first equation.
Entries from NumNodes+1 to 2*NumNodes correspond to the second equation.
Entries from 2*NumNodes+1 to 3*NumNodes correspond to the third equation.
The same approach applies to all other entries, up to N*NumNodes.
For example, in a 3-D structural model, the length of a solution vector u is 3*NumNodes. The first NumNodes entries correspond to the x-displacement at each node, the next NumNodes entries correspond to the y-displacement, and the next NumNodes entries correspond to the z-displacement.
In thermal analysis, the m and a coefficients are zeros. The thermal conductivity maps to the c coefficient. The product of the mass density and the specific heat maps to the d coefficient. The internal heat source maps to the f coefficient. The temperature on a boundary corresponds to the Dirichlet boundary condition term r with h = 1. Various forms of boundary heat flux, such as the heat flux itself, emissivity, and convection coefficient, map to the Neumann boundary condition terms q and g.
In structural analysis, the a coefficient is zero. Young's modulus and Poisson's ratio map to the c coefficient. The mass density maps to the m coefficient. The body loads map to the f coefficient. Displacements, constraints, and components of displacement along the axes map to the Dirichlet boundary condition terms h and r. Boundary loads, such as pressure, surface tractions, and translational stiffnesses, correspond to the Neumann boundary condition terms q and g. When you specify the damping model by using the Rayleigh damping parameters Alpha and Beta, the discretized damping matrix C is computed by using the mass matrix M and the stiffness matrix K as C = Alpha*M + Beta*K. Hysteretic (structural) damping contributes to the stiffness matrix K, which becomes complex.
In electrostatic and magnetostatic analyses, the m, a, and d coefficients are zeros. The relative permittivity and relative permeability map to the c coefficient. The charge density and current density map to the f coefficient. The voltage and magnetic potential on a boundary correspond to the Dirichlet boundary condition term r with h = 1.
Assembling FE matrices does not work for harmonic analysis and 3-D magnetostatic analysis.
PDEModel | ThermalModel | StructuralModel | ElectromagneticModel | solvepde | solve
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(Redirected from JJ thomson)
Thomson was awarded the 1906 Nobel Prize in Physics for his work on the conduction of electricity in gases.[4] Thomson was also a teacher, and nine of his mentees also went on to win Nobel Prizes.[citation needed]
{\displaystyle \Theta =Fel/mv^{2}}
{\displaystyle \phi =Hel/mv}
{\displaystyle \Theta =\phi ,Fel/mv^{2}=Hel/mv}
{\displaystyle m/e=H^{2}l/F\Theta }
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General_position Knowpia
General linear positionEdit
A set of points in a d-dimensional affine space (d-dimensional Euclidean space is a common example) is in general linear position (or just general position) if no k of them lie in a (k − 2)-dimensional flat for k = 2, 3, ..., d + 1. These conditions contain considerable redundancy since, if the condition holds for some value k0 then it also must hold for all k with 2 ≤ k ≤ k0. Thus, for a set containing at least d + 1 points in d-dimensional affine space to be in general position, it suffices that no hyperplane contains more than d points — i.e. the points do not satisfy any more linear relations than they must.[1]
A set of at most d + 1 points in general linear position is also said to be affinely independent (this is the affine analog of linear independence of vectors, or more precisely of maximal rank), and d + 1 points in general linear position in affine d-space are an affine basis. See affine transformation for more.
Similarly, n vectors in an n-dimensional vector space are linearly independent if and only if the points they define in projective space (of dimension n − 1) are in general linear position.
More generallyEdit
This is not sufficient, however. While nine points determine a cubic, there are configurations of nine points that are special with respect to cubics, namely the intersection of two cubics. The intersection of two cubics, which is
{\displaystyle 3\times 3=9}
points (by Bézout's theorem), is special in that nine points in general position are contained in a unique cubic, while if they are contained in two cubics they in fact are contained in a pencil (1-parameter linear system) of cubics, whose equations are the projective linear combinations of the equations for the two cubics. Thus such sets of points impose one fewer condition on cubics containing them than expected, and accordingly satisfy an additional constraint, namely the Cayley–Bacharach theorem that any cubic that contains eight of the points necessarily contains the ninth. Analogous statements hold for higher degree.
Different geometriesEdit
General typeEdit
General position is a property of configurations of points, or more generally other subvarieties (lines in general position, so no three concurrent, and the like). General position is an extrinsic notion, which depends on an embedding as a subvariety. Informally, subvarieties are in general position if they cannot be described more simply than others. An intrinsic analog of general position is general type, and corresponds to a variety which cannot be described by simpler polynomial equations than others. This is formalized by the notion of Kodaira dimension of a variety, and by this measure projective spaces are the most special varieties, though there are other equally special ones, meaning having negative Kodaira dimension. For algebraic curves, the resulting classification is: projective line, torus, higher genus surfaces (
{\displaystyle g\geq 2}
), and similar classifications occur in higher dimensions, notably the Enriques–Kodaira classification of algebraic surfaces.
General position for Delaunay triangulations in the planeEdit
When discussing Voronoi tessellations and Delaunay triangulations in the plane, a set of points in the plane is said to be in general position only if no four of them lie on the same circle and no three of them are collinear. The usual lifting transform that relates the Delaunay triangulation to the bottom half of a convex hull (i.e., giving each point p an extra coordinate equal to |p|2) shows the connection to the planar view: Four points lie on a circle or three of them are collinear exactly when their lifted counterparts are not in general linear position.
Abstractly: configuration spacesEdit
^ Yale 1968, p. 164
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Abstract: A unified field theory is created by using a simple dipole antenna to model the Earth’s electrical fields. From Maxwell’s equation a set of standard radio field equations are presented for the Electrostatic, Inductive and Radiative field. By inserting three gravity field equations (1/r, 1/r2 and 1/r3) a new set of field equations are created that can be used to define space, time, gravity and the electromagnetic field at any point P in the field.
Keywords: Dipole Antenna, Gravity Field, Maxwell’s Equation, Unified Field Theory
In this paper, I will use simple dipole radio equations, based on Maxwell equations, and insert my newly discovered gravity field equations for the 1/r, 1/r2 and 1/r3 electromagnetic fields. By adding the contribution of each field, total gravity is achieved. The final series of equations will then include space, time, gravity and electromagnetism, which can be used to identify the variable at any point P in the fields. A simple antenna is thus used as a means to develop a unified field theory.
2. History of Unified Field Theory
In 1864, James Clerk Maxwell published his famous paper on a dynamical theory of the electromagnetic field. This was the first example of a theory that was able to encompass previously separate electrostatic and magnetic field theories to provide a unifying theory of electromagnetism. By 1905, Albert Einstein had used the constancy of the speed of light in Maxwell’s theory to unify our notions of space and time into an entity we now call spacetime. In 1915, Einstein expanded this theory of special relativity to general relativity, using a gravity field to describe the curving geometry of four-dimensional spacetime. He created his gravity field from a centrifugal field using an equivalency principle (Figure 1).
Figure 1. Radio fields & gravity decay over distance.
The term Unified Field Theory was coined by Albert Einstein [1] , who attempted to unify his general theory of relativity with electromagnetism. The “Theory of Everything” [2] and Grand Unified Theory [3] are closely related to unified field theory, but differ by not requiring the basis of nature to be fields, and often by attempting to explain physical constants of nature. Einstein relativity work is based on a unipolar (monopole) electrodynamic model, as opposed to my dipole model.
Following the creation of the general relativity theory, a large number of physicists and mathematicians attempted to unify the then-known fundamental interactions [4] . A large number of mathematicians and physicists, including Hermann Weyl, Arthur Eddington, and Theodor Kaluza also attempted to develop approaches that could unify these interactions [5] [6] . These scientists pursued several avenues of generalization, including extending the foundations of geometry and adding an extra spatial dimension.
In view of later developments in this domain, of particular interest are the theories of Hermann Weyl from 1919, who introduced the concept of an (electromagnetic) gauge field in a classical field theory [7] and, two years later, that of Theodor Kaluza, who extended General Relativity to five dimensions [8] . Continuing in this latter direction, Oscar Klein, in 1926, proposed that the fourth spatial dimension be curled up into a small, unobserved circle. In Kaluza-Klein theory, the gravitational curvature of the extra spatial direction behaves as an additional force similar to electromagnetism. These and other models of electromagnetism and gravity were pursued by Albert Einstein in his attempts at a classical unified field theory. By 1930, Einstein had already considered the Einstein-Maxwell-Dirac System.
3. Gravity Field Equations
Gravity Far Field:
{g}_{F}=\frac{G{c}^{2}}{r}
{c}^{2}r=\frac{{g}_{F}{r}^{2}}{G}
Gravity Near Field:
{g}_{N}=\frac{G{c}^{3}}{2\text{π}{r}^{2}}
c{r}^{2}=\frac{2\text{π}{g}_{N}{r}^{4}}{G{c}^{2}}
Gravity Magnetoquasistatic (Electrostatic) Field:
{g}_{M}=\frac{G{c}^{4}\tau }{\text{π}{r}^{3}}
\tau =1/f
\omega =2\text{π}f=2\text{π}/\tau
Inserting, I obtain
\tau =\frac{{g}_{M}\text{π}{r}^{3}}{G{c}^{4}}
\frac{2\text{π}}{\omega }=\frac{{g}_{M}\text{π}{r}^{3}}{G{c}^{4}}
\omega {r}^{3}=\frac{2G{c}^{4}}{{g}_{M}}
4. Field Equations of an Antenna
From Maxwell’s equations the components of the fields at any point P are:
{H}_{\varphi }=Il\mathrm{sin}\theta \left[-\frac{\mathrm{cos}\omega \left(t-\frac{r}{c}\right)}{c}\frac{1}{{r}^{2}}+\frac{\omega \mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{{c}^{2}}\frac{1}{r}\right]
{H}_{r}=0
{H}_{\theta }=0
{E}_{\varphi }=0
{E}_{r}=2Il\mathrm{cos}\theta \left[\frac{\mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{\omega }\frac{1}{{r}^{3}}+\frac{\mathrm{cos}\omega \left(t-\frac{r}{c}\right)}{c}\frac{1}{{r}^{2}}\right]
{E}_{\theta }=Il\mathrm{sin}\theta \left[\frac{\mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{\omega }\frac{1}{{r}^{3}}+\frac{\mathrm{cos}\omega \left(t-\frac{r}{c}\right)}{c}\frac{1}{{r}^{2}}+\frac{\omega \mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{{c}^{2}}\frac{1}{r}\right]
r\gg \lambda
we have the radiation fields of the antenna:
{H}_{\varphi }=Il\mathrm{sin}\theta \left[\frac{\omega \mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{{c}^{2}}\frac{1}{r}\right]
{E}_{r}=0
{E}_{\theta }=Il\mathrm{sin}\theta \left[\frac{\omega \mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{{c}^{2}}\frac{1}{r}\right]
5. Electro Relativity Equations
{H}_{\varphi }=Il\mathrm{sin}\theta \left[-\frac{G{c}^{2}\mathrm{cos}\omega \left(t-\frac{r}{c}\right)}{2\text{π}{g}_{N}{r}^{4}}+\frac{G\omega \mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{{g}_{F}{r}^{2}}\right]
{H}_{r}=0
{H}_{\theta }=0
{E}_{\varphi }=0
{E}_{r}=2Il\mathrm{cos}\theta \left[\frac{{g}_{M}\mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{2G{c}^{4}}+\frac{G{c}^{2}\mathrm{cos}\omega \left(t-\frac{r}{c}\right)}{2\text{π}{g}_{N}{r}^{4}}\right]
{E}_{\theta }=Il\mathrm{sin}\theta \left[\frac{{g}_{M}\mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{2G{c}^{4}}+\frac{G{c}^{2}\mathrm{cos}\omega \left(t-\frac{r}{c}\right)}{2\text{π}{g}_{N}{r}^{4}}+\frac{G\omega \mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{{g}_{F}{r}^{2}}\right]
r\gg \lambda
{H}_{\varphi }=Il\mathrm{sin}\theta \left[\frac{G\omega \mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{{g}_{F}{r}^{2}}\right]
{E}_{r}=0
{E}_{\theta }=Il\mathrm{sin}\theta \left[\frac{G\omega \mathrm{sin}\omega \left(t-\frac{r}{c}\right)}{{g}_{F}{r}^{2}}\right]
Modeling of the Earth as a simple dipole antenna provides us with a view of the Earth’s three-dimensional spherical electromagnetic field. Using the speed of light (c), the radius of the Earth (or field), Newton’s Constant (G) for proportionating and the flux transfer event frequency (f), a new approach to gravity is explored using electromagnetism. By inserting three gravity field equations into standard radio equations a unified field theory is created that describes space, time, gravity and electromagnetism at any point P in the field. Gravity at any point P is determined by
{g}_{T}={g}_{M}+{g}_{N}+{g}_{F}
We can readily conclude that maximum acceleration of light and compression of space in the magnetoquasistatic field give us a solid planet. And, atmospheric pressure of gases is derived from somewhat less acceleration and compression of space in the near field. We also know that atmospheric pressure has a mathematical relationship to gravity. No compression of space occurs in the far field.
The author wishes to acknowledge ASK Scientific for assistance in formatting the equations. The work of James Clerk Maxwell, Sir Joseph Larmor, Heinrick Hertz, Nickola Tesla and Guglielmo Marconi, the inventors of radio were instrumental in creating this electro relativity theory.
Cite this paper: Poole, G. (2019) Theory of Electro Relativity. Journal of High Energy Physics, Gravitation and Cosmology, 5, 1063-1067. doi: 10.4236/jhepgc.2019.54059.
[1] How the Search for a Unified Theory Stumped Einstein to His Dying Day. phys.org.
[2] Hawking, S.W. (2006) The Theory of Everything: The Origin and Fate of the Universe. Phoenix Books.
[3] Ross, G. (1984) Grand Unified Theories. Westview Press, Boulder.
[4] Vizgin, V. (1994) Unified Field Theories in the First Third of the 20th Century. Birkhäuser, Basel.
[5] Weyl, H. (1918) Gravitation und Elektrizität. Sitzungsber. Preuss. Akad. Wiss., Berlin, 465.
[6] Eddington, A.S. (1924) The Mathematical Theory of Relativity. 2nd Edition, Cambridge Univ. Press, Cambridge.
[7] Scholtz, E. (2001) Hermann Weyl’s Raum—Zeit—Materie and a General Introduction to His Scientific Work. Birkhäuser, Basel.
[8] Wuensch, D. (2003) The Fifth Dimension: Theodor Kaluza’s Ground-Breaking Idea. Annalen der Physik, 12, 519-542.
[9] Electrical Engineering of the Westinghouse Electric Corporation (1948) Industrial Electronics Reference Book. Chapter 20, John Wiley & Sons, Inc., Hoboken, 308-310.
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Talatuntunan ng kulay - Wikipedia, ang malayang ensiklopedya
Talatuntunan ng kulay
Sampol ng mga kalibrasyon ng kulay[1]
Sa astronomiya, ang talatuntunan ng kulay (color index) ay isang payak na ekspresyong numerikal na nagdedetermina sa kulay ng isang bagay, na sa kaso ng bituin ay ang kanilang temperatura. Para masukat ang talatuntunan, tuluyang inoobserbahan ang magnitud ng isang bagay gamit ang dalawang magkaibang pansala sa sistemang UBV, tulad ng U at B, o B at V, kung saan sensitibo ang U sa mga silahis na ultrabiyoleta, sensitibo ang B sa bughaw na liwanag, at sensitibo ang V sa liwanag na makikita (luntian-dilaw). The set of passbands or filters is called a photometric system. The difference in magnitudes found with these filters is called the U-B or B–V color index, respectively. The smaller the color index, the more blue (or hotter) the object is. Conversely, the larger the color index, the more red (or cooler) the object is. This is a consequence of the logarithmic magnitude scale, in which brighter objects have smaller (more negative) magnitudes than dimmer ones. For comparison, the yellowish Sun has a B–V index of 0.656 ± 0.005,[2] while the bluish Rigel has B–V –0.03 (its B magnitude is 0.09 and its V magnitude is 0.12, B–V = –0.03).[3]
Color indices of distant objects are usually affected by interstellar extinction —i.e. they are redder than those of closer stars. The amount of reddening is characterized by color excess, defined as the difference between the Observed color index and the Normal color index (or Intrinsic color index), the hypothetical true color index of the star, unaffected by extinction. For example, in the UBV photometric system we can write it for the B-V color:
{\displaystyle E_{B-V}=(B-V)_{\textrm {Observed}}-(B-V)_{\textrm {Intrinsic}}}
The passbands most optical astronomers use are the UBVRI filters, where the U, B, and V filters are as mentioned above, the R filter passes red light, and the I filter passes infrared light. This system of filters is sometimes called the Johnson-Cousins filter system, named after the originators of the system (see references). These filters were specified as particular combinations of glass filters and photomultiplier tubes. M. S. Bessell specified a set of filter transmissions for a flat response detector, thus quantifying the calculation of the color indices.[4] For precision, appropriate pairs of filters are chosen depending on the object's color temperature: B-V are for mid-range objects, U-V for hotter objects, and R-I for cool ones.
↑ Zombeck, Martin V. (1990). "Calibration of MK spectral types". Handbook of Space Astronomy and Astrophysics (ika-2nd (na) edisyon). Cambridge University Press. pa. 105. ISBN 0-521-34787-4.
↑ Michael S. Bessell (1990), UBVRI passbands, Astronomical Society of the Pacific, Publications (ISSN 0004-6280), vol. 102, Oct. 1990, p. 1181-1199
Kinuha sa "https://tl.wikipedia.org/w/index.php?title=Talatuntunan_ng_kulay&oldid=1350603"
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decorate - Simple English Wiktionary
IPA (key): /ˈdɛkəreɪt/
A decorated vehicle
(transitive & intransitive) If you decorate something, you make it look nice, usually by putting something pretty on it or painting it.
We're going to decorate our Christmas tree tonight.
The men sat in large uncomfortable wooden chairs decorated with carvings of flowers.
The room was brightly decorated.
{\displaystyle x}
{\displaystyle y}
{\displaystyle y}
looks nice because
{\displaystyle x}
is on or close to it.
In her room, photos decorated the white walls.
(transitive) If you decorate someone, you give them a medal for doing something brave or good.
He was decorated by the president for saving several men who had been shot.
Retrieved from "https://simple.wiktionary.org/w/index.php?title=decorate&oldid=484309"
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(Redirected from Universal transverse Mercator)
4.1 Simplified formulae
The WGS 84 spatial reference system describes Earth as an oblate spheroid along north-south axis with an equatorial radius of
{\displaystyle a=6378.137}
km and an inverse flattening of
{\displaystyle 1/f=298.257\,223\,563}
. Let's take a point of latitude
{\displaystyle \,\varphi }
and of longitude
{\displaystyle \,\lambda }
and compute its UTM coordinates as well as point scale factor
{\displaystyle k\,\!}
and meridian convergence
{\displaystyle \gamma \,\!}
using a reference meridian of longitude
{\displaystyle \lambda _{0}}
. By convention, in the northern hemisphere
{\displaystyle N_{0}=0}
km and in the southern hemisphere
{\displaystyle N_{0}=10000}
km. By convention also
{\displaystyle k_{0}=0.9996}
{\displaystyle E_{0}=500}
{\displaystyle n={\frac {f}{2-f}},\quad A={\frac {a}{1+n}}\left(1+{\frac {n^{2}}{4}}+{\frac {n^{4}}{64}}+\cdots \right),}
{\displaystyle \alpha _{1}={\frac {1}{2}}n-{\frac {2}{3}}n^{2}+{\frac {5}{16}}n^{3},\,\,\,\alpha _{2}={\frac {13}{48}}n^{2}-{\frac {3}{5}}n^{3},\,\,\,\alpha _{3}={\frac {61}{240}}n^{3},}
{\displaystyle \beta _{1}={\frac {1}{2}}n-{\frac {2}{3}}n^{2}+{\frac {37}{96}}n^{3},\,\,\,\beta _{2}={\frac {1}{48}}n^{2}+{\frac {1}{15}}n^{3},\,\,\,\beta _{3}={\frac {17}{480}}n^{3},}
{\displaystyle \delta _{1}=2n-{\frac {2}{3}}n^{2}-2n^{3},\,\,\,\delta _{2}={\frac {7}{3}}n^{2}-{\frac {8}{5}}n^{3},\,\,\,\delta _{3}={\frac {56}{15}}n^{3}.}
{\displaystyle t=\sinh \left(\tanh ^{-1}\left(\sin \varphi \right)-{\frac {2{\sqrt {n}}}{1+n}}\tanh ^{-1}\left({\frac {2{\sqrt {n}}}{1+n}}\sin \varphi \right)\right),}
{\displaystyle \xi '=\tan ^{-1}\left({\frac {t}{\cos(\lambda -\lambda _{0})}}\right),\,\,\,\eta '=\tanh ^{-1}\left({\frac {\sin(\lambda -\lambda _{0})}{\sqrt {1+t^{2}}}}\right),}
{\displaystyle \sigma =1+\sum _{j=1}^{3}2j\alpha _{j}\cos(2j\xi ')\cosh(2j\eta '),\,\,\,\tau =\sum _{j=1}^{3}2j\alpha _{j}\sin(2j\xi ')\sinh(2j\eta ').}
{\displaystyle E=E_{0}+k_{0}A\left(\eta '+\sum _{j=1}^{3}\alpha _{j}\cos(2j\xi ')\sinh(2j\eta ')\right),}
{\displaystyle N=N_{0}+k_{0}A\left(\xi '+\sum _{j=1}^{3}\alpha _{j}\sin(2j\xi ')\cosh(2j\eta ')\right),}
{\displaystyle k={\frac {k_{0}A}{a}}{\sqrt {\left\{1+\left({\frac {1-n}{1+n}}\tan \varphi \right)^{2}\right\}{\frac {\sigma ^{2}+\tau ^{2}}{t^{2}+\cos ^{2}(\lambda -\lambda _{0})}}}},}
{\displaystyle \gamma =\tan ^{-1}\left({\frac {\tau {\sqrt {1+t^{2}}}+\sigma t\tan(\lambda -\lambda _{0})}{\sigma {\sqrt {1+t^{2}}}-\tau t\tan(\lambda -\lambda _{0})}}\right).}
{\displaystyle E}
is Easting,
{\displaystyle N}
is Northing,
{\displaystyle k}
is the Scale Factor, and
{\displaystyle \gamma }
{\displaystyle \xi ={\frac {N-N_{0}}{k_{0}A}},\,\,\,\eta ={\frac {E-E_{0}}{k_{0}A}},}
{\displaystyle \xi '=\xi -\sum _{j=1}^{3}\beta _{j}\sin \left(2j\xi \right)\cosh \left(2j\eta \right),\,\,\,\eta '=\eta -\sum _{j=1}^{3}\beta _{j}\cos \left(2j\xi \right)\sinh \left(2j\eta \right),}
{\displaystyle \sigma '=1-\sum _{j=1}^{3}2j\beta _{j}\cos \left(2j\xi \right)\cosh \left(2j\eta \right),\,\,\,\tau '=\sum _{j=1}^{3}2j\beta _{j}\sin \left(2j\xi \right)\sinh \left(2j\eta \right),}
{\displaystyle \chi =\sin ^{-1}\left({\frac {\sin \xi '}{\cosh \eta '}}\right).}
{\displaystyle \varphi =\chi +\sum _{j=1}^{3}\delta _{j}\sin \left(2j\chi \right),}
{\displaystyle \lambda _{0}=\mathrm {Z} \mathrm {o} \mathrm {n} \mathrm {e} \times 6^{\circ }-183^{\circ }\,}
{\displaystyle \lambda =\lambda _{0}+\tan ^{-1}\left({\frac {\sinh \eta '}{\cos \xi '}}\right),}
{\displaystyle k={\frac {k_{0}A}{a}}{\sqrt {\left\{1+\left({\frac {1-n}{1+n}}\tan \varphi \right)^{2}\right\}{\frac {\cos ^{2}\xi '+\sinh ^{2}\eta '}{\sigma '^{2}+\tau '^{2}}}}},}
{\displaystyle \gamma =\mathrm {H} \mathrm {e} \mathrm {m} \mathrm {i} \times \tan ^{-1}\left({\frac {\tau '+\sigma '\tan \xi '\tanh \eta '}{\sigma '-\tau '\tan \xi '\tanh \eta '}}\right).}
Retrieved from "https://en.wikipedia.org/w/index.php?title=Universal_Transverse_Mercator_coordinate_system&oldid=1081926010"
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What is the tangent ratio in trigonometry? | StudyPug
\frac{o}{a}
\frac{o}{a}
Tangent ratio is yet another trigonometric ratio for right-angled triangles. Tangent ratio is the ratio of opposite side to adjacent side of a right triangle. Same as the sine and cosine ratios, tangent ratios can be used to calculate the angles and sides of right angle triangles.
Related Concepts: Law of sines, Law of cosines, Find the exact value of trigonometric ratios, Unit circle
Determine each tangent ratio using a calculator
\tan 70^\circ
\tan-134^\circ
Determine the angle to the nearest degree using a a calculator
\tan \theta = 0.64
\tan \theta = -1.45
Determine the angles and sides using Tangent
Find the value of "x" using tangent
Draw and label a right triangle to illustrate tangent ratio, then calculate the angle.
\tan \theta = {5 \over 9}
\frac{o}{h}
\frac{a}{h}
\frac{o}{a}
\frac{o}{a}
\frac{o}{h}
\frac{a}{h}
\frac{o}{a}
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Revision as of 11:52, 9 April 2017 by MathAdmin (talk | contribs) (Created page with "<span class="exam">Find the derivatives of the following functions. Do not simplify. <span class="exam">(a) <math style="vertical-align: -5px">f(x)=x^3(x^{\frac{4}{3}}...")
{\displaystyle f(x)=x^{3}(x^{\frac {4}{3}}-1)}
{\displaystyle g(x)={\frac {x^{3}+x^{-3}}{1+6x}}}
{\displaystyle x>0}
{\displaystyle {\frac {d}{dx}}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)}
{\displaystyle {\frac {d}{dx}}{\bigg (}{\frac {f(x)}{g(x)}}{\bigg )}={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}}
{\displaystyle {\frac {d}{dx}}(x^{n})=nx^{n-1}}
{\displaystyle f'(x)=x^{3}(x^{\frac {4}{3}}-1)'+(x^{3})'(x^{\frac {4}{3}}-1).}
{\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {x^{3}(x^{\frac {4}{3}}-1)'+(x^{3})'(x^{\frac {4}{3}}-1)}\\&&\\&=&\displaystyle {x^{3}{\bigg (}{\frac {4}{3}}x^{\frac {1}{3}}{\bigg )}+(3x^{2})(x^{\frac {4}{3}}-1).}\end{array}}}
{\displaystyle g'(x)={\frac {(1+6x)(x^{3}+x^{-3})'-(x^{3}+x^{-3})(1+6x)'}{(1+6x)^{2}}}.}
{\displaystyle {\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {\frac {(1+6x)(x^{3}+x^{-3})'-(x^{3}+x^{-3})(1+6x)'}{(1+6x)^{2}}}\\&&\\&=&\displaystyle {{\frac {(1+6x)(3x^{2}-3x^{-4})-(x^{3}+x^{-3})(6)}{(1+6x)^{2}}}.}\end{array}}}
{\displaystyle x^{3}{\bigg (}{\frac {4}{3}}x^{\frac {1}{3}}{\bigg )}+(3x^{2})(x^{\frac {4}{3}}-1)}
{\displaystyle {\frac {(1+6x)(3x^{2}-3x^{-4})-(x^{3}+x^{-3})(6)}{(1+6x)^{2}}}}
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The command CellPlot(
m
k
) plots the
k
th cell of the cell decomposition given by the solution module
m
k
is a list, then all cells whose indices are in
k
are plotted. If
k
is omitted entirely, then all open cells are plotted.
Currently, this command can only be used in the case of exactly
2
parameters. A corresponding error message is returned otherwise.
thickness=t, where t is a positive integer (the default is
2
), controls the thickness of the boundary of each cell.
\mathrm{with}\left(\mathrm{RootFinding}[\mathrm{Parametric}]\right):
m≔\mathrm{CellDecomposition}\left([{x}^{3}+a{x}^{2}+bx+a=0],[x],[a,b]\right):
\mathrm{CellPlot}\left(m\right)
\mathrm{CellDescription}\left(m,2\right)
[[\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{\infty }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{a}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{a}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{b}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{18}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{a}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{b}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{27}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{a}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{a}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{a}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{b}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{18}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{a}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{b}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{27}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{a}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}]]
\mathrm{CellPlot}\left(m,2,\mathrm{color}=\mathrm{yellow}\right)
\mathrm{CellPlot}\left(m,[1,2,3,4],\mathrm{samplepoints}=\mathrm{true},\mathrm{pointcolor}=\mathrm{magenta},\mathrm{view}=[-2..2,-2..2]\right)
m≔\mathrm{CellDecomposition}\left([{x}^{3}+a{x}^{2}+bx+a=0],[x],[b,a]\right):
\mathrm{CellPlot}\left(m\right)
\mathrm{CellPlot}\left(m,[2,3],\mathrm{cellcolor}=[\mathrm{blue},\mathrm{red}]\right)
\mathrm{with}\left(\mathrm{RootFinding}\right):
\mathrm{EnclosingBox}\left(m\right):
\mathrm{CellPlot}\left(m,\mathrm{samplepoints}\right)
|
Introduction to Chemical Engineering Processes/Excel - Wikibooks, open books for an open world
2 Anatomy of a spreadsheet
3 Inputting and Manipulating Data in Excel
3.1 Using formulas
3.2 Performing Operations on Groups of Cells
3.3 Special Functions in Excel
3.3.1 Mathematics Functions
3.3.2 Statistics Functions
3.3.3 Programming Functions
4 Solving Equations in Spreadsheets: Goal Seek
5 Graphing Data in Excel
5.2 Performing Regressions of the Data from a Scatterplot
6 Further resources for Spreadsheets
Introduction to SpreadsheetsEdit
This tutorial probably works with other spreadsheets (such as w:open office) with minor modifications.
A spreadsheet such as Excel is a program that lets you analyze moderately large amounts of data by placing each data point in a cell and then performing the same operation on groups of cells at once. One of the nice things about spreadsheets is that data input and manipulation is relatively intuitive and hence easier than doing the same tasks in a programming language like MATLAB (discussed next). This section shows how to do some of these manipulations so that you don't have to by hand.
Anatomy of a spreadsheetEdit
A spreadsheet has a number of parts that you should be familiar with. When you first open up the spreadsheet program, you will see something that looks like this (the image is from the German version of open office)
First off, notice that the entire page is split up into boxes, and each one is labeled. Rows are labeled with numbers and columns with letters. Also, try typing something in, and notice that the box above the spreadsheet (to the right of
{\displaystyle \Sigma =}
) will change automatically as you type. When you're just putting in numbers, this info box will just have the same number in there. But when you're putting in formulas, the cell will display the value calculated from the formula, while the info box will display what the formula was.
Inputting and Manipulating Data in ExcelEdit
The first step in any spreadsheet analysis is to input the raw data you want to analyze. It is most effective if you put it in columns, with one column for each variable. It lets you see more data at once, and it also is less limited because the maximum number of rows is much larger than the maximum number of columns.
It is good practice to use the first row for the names of the variables, and the remaining for the data points. Make sure you include units. In this section, the following data will be used as illustration:
(Rownumber)
1 t(min) D (yards)
Using formulasEdit
In order to tell the spreadsheet that you want to use a formula rather than just enter a number, you have to start the entry with an equal sign (=). You can then use combinations of decimal values and cell designations. A cell designation is simply the column letter followed by the row number containing the value you wish to manipulate. For example, if you wanted to find the product of the distance traveled and the time spent traveling, you could put in the formula:
into any empty cell and it would give you the answer. From here out it will be assumed that this value is in cell C2. You should label the column with the type of calculation you're performing.
Performing Operations on Groups of CellsEdit
The question may arise: why not just put in the numbers themselves instead of referencing the cell? There are two major reasons for this:
If you change the value in the referenced cell, the value calculated in the formula will automatically change.
The built-in dragging capability of most spreadsheets.
The dragging capability is a simple concept. If you have put a formula into a spreadsheet, you can have it copied to any number of cells you want. To do this, select the cell with the formula and bring the mouse pointer to its lower-right hand corner. You should see a dark + icon:
Info Bar "=A2*B2"
(rownum) Column A Column B Column C
1 t(min) D(yards) t*D
Click on the + and drag it down. This will cause the formulas to change according to how you drag the box. In this case, if you drag it down to row 6, the spreadsheet will produce the following:
A B C
1 t (min) D (yards) t*D
2 1.1 559.5 615.45
3 1.9 759.5 1443.05
4 3.0 898.2 2694.6
5 3.8 1116.3 4241.94
If you click on the last value in column C (6936.11) the info bar will display:
This is very useful for performing the same operations on multiple sets of data at once; rather than having to do the multiplication 5 separate times here, we just do it once and drag down the box.
Special Functions in ExcelEdit
In order to do many mathematical operations in Excel (or at least the easiest way), it is necessary to use functions (not to be confused with formulas). A function is simply an implementation someone already wrote for the mathematical operation, so all you have to do is know how to tell it to do the operation and where to put it when it's done. In excel, you can call a function named "function" by typing the following into a cell:
=function(inputs)
The function will then execute, and the cell containing the call will display the answer. The necessary inputs are sometimes numbers but are more often the cell addresses. For example, in the data above, say you wanted to take the exponential (
{\displaystyle e^{x}}
) of all the time points in column A, and place the result in column D. The function for exponential is exp, and it can only accept one input at a time, but due to the dragging capability of Excel this will not matter much, you can just call it once and then drag the cell as you would with any formula containing cell addresses. So to do this you would type into cell D2:
Hit enter, then click the + in the bottom right and drag the cell down. You should end up with something like this after labeling the D column appropriately:
A B C D
1 t (min) D(yards) t*D e^t
2 1.1 559.5 615.45 3.004166024
3 1.9 759.5 1443.05 6.685894442
4 3 898.2 2694.6 20.08553692
5 3.8 1116.3 4241.94 44.70118449
6 5.3 1308.7 6936.11 200.33681
All excel functions output only one value at a time, though some can accept multiple cells at a time as input (mostly statistical functions).
Following is a brief synopsis of the functions available. For a complete list, see the help files for your spreadsheet, as the availability of each function may vary depending on which one you are using. CELL signifies either the row/column designation of the cell you want to pass to the function as input, or some numerical value you enter manually.
Mathematics FunctionsEdit
Generally these only take one input at a time.
abs(CELL): Absolute value of CELL
sqrt(CELL): Square root of CELL [to do nth roots, use CELL^(1/n)]
ln(CELL): Natural log of CELL
log10(CELL): Log of CELL to base 10
log(CELL, NUM): Log of CELL to the base NUM (use for all bases except e and 10)
exp(CELL): Exponential(e^x) of CELL. Use since Excel doesn't have a built-in constant "e".
sin(CELL), cos(CELL), tan(CELL): Trigonometric functions sine, cosine, and tangent of CELL. CELL must be in radians
asin(CELL), acos(CELL), atan(CELL): Inverse trigonometric functions (returns values in radians)
sinh(CELL), cosh(CELL), tanh(CELL): Hyperbolic functions
asinh(CELL), acosh(CELL), atanh(CELL): Inverse hyperbolic functions
Statistics FunctionsEdit
These are examples of useful statistics functions in Excel, they are not by any means the only ones.
GROUP means a group of cells that are directly next to each other. Define a group by the syntax FIRSTCELL:LASTCELL, for example, using GROUP = A2:A5 passes all the cells between A2 and A5 (inclusive) to the function. If a function requires two different groups (for example, a y and an x), both groups must be within continuous groups of cells.
average(CELL1, CELL2, ...) OR average(GROUP): Computes the arithmetic average of all inputs.
intercept(GROUP1, GROUP2): Calculates the y-intercept (b) of the regression line where y = GROUP1 and x = GROUP2.
GROUP1 and GROUP2 must have the same size.
pearson(GROUP1, GROUP2): Calculates the Pearson correlation coefficient (R) between GROUP1 and GROUP2.
stdev(CELL1, CELL2, ...) OR stdev(GROUP): Computes the sample standard deviation (divides by n-1) of all inputs.
slope(GROUP1, GROUP2): Calculates the slope (m) of the regression line where y = GROUP1 and x = GROUP2.
Programming FunctionsEdit
Solving Equations in Spreadsheets: Goal SeekEdit
Excel and possibly other spreadsheets have a very useful tool called goalseek which allows the user to solve single-variable equations (and can be used as an aid in guess-and-check for systems of algebraic equations). Let's suppose for the purposes of this tutorial that you wish to find a solution to the equation:
{\displaystyle 0=X^{3}+2X^{2}-X+1}
In order to set up the problem in Goalseek, it is necessary to define a cell for the variable you want to change (X) and a cell for the function you want to evaluate.
Goalseek will only work if you tell it to evaluate some function until it reaches a constant value. You cannot tell it to equal something that can change, so for example you cannot do something like this:
{\displaystyle X-X^{3}=2X^{2}+1}
because neither side is a constant. The easiest way around this is generally to solve the function for zero and then use that as the evaluating function.
Here, we could set up the cells as follows:
1 X f(X)
2 -1 =A2^3 + 2*A2^2 - A2 + 1
To solve this one, go to
Tools > Goalseek...
It'll give you three boxes: "Set Cell", "To Value", and "By Changing Cell". Since we want the value in cell B2 to equal 0, enter B2 into the "Set Cell" box and 0 into the "To Value" box. Since cell B2 depends on cell A2, we want to change A2 so that B2 equals 0. Hence, the "By Changing Cell" box should contain A2. Put that in and click "OK", and Goalseek will converge to an answer:
1 X f(X)
2 -2.54683 -0.00013
Notice that the success of the goalseek depends on what your initial guess was. If you try to put in an initial guess of 0 in this example (instead of -1), goalseek will diverge. It will tell you so, saying "Goal Seeking with Cell B2 May Not have Found a Solution". However, the algorithm is generally fairly robust so it shouldn't take too many guesses to obtain convergence.
You can only enter one cell into each of the "Set Cell" and "By Changing Cell" boxes, and the value in "To Value" must be a constant
Graphing Data in ExcelEdit
In Excel, there are a variety of ways to graph the data you have inserted, such as bar graphs, pie charts, and many others. The most commonly-used in my experience is the scatterplot, which is the name Excel uses for the typical x-y "line graph" plot that you probably think of first when you think of a graph.
ScatterplotsEdit
Scatterplots can be made relating any one independent variable to any number of dependent variables, though if you try to graph too many it will get crowded and hard to read. Excel will automatically give each different dependent variable a different color and a different shape, so that you can distinguish between them. You can also name each "series" of data differently and Excel will automatically set up a legend for you.
This is how to make a scatterplot:
Put the data into columns just like it was given in the problem statement.
Now we need to set up the graph. Go to:
{\displaystyle Insert\rightarrow Chart}
Select "XY (scatter)" and click "next".
Click the "series" tab (on top). If there are any series present, remove them with the remove button (since it usually guesses wrong what you want to graph).
Now we can add a series for each dependent variable we want to graph as follows:
Next to "X values" click the funky arrow symbol to the right of the text box. A small box will pop up.
Click on the first value for the independent variable and drag the mouse down to the last value. Click the funky symbol again to bring you back to the main window.
Do the same thing with the "Y values" but this time you want to select the values of the dependent variable.
Click next, and give the graph a title and labels if you want. Then click next and "finish" to generate your graph.
Performing Regressions of the Data from a ScatterplotEdit
Once you have a scatterplot of your data, you can do one of several types of regression: logarithmic, exponential, polynomial (up to 6th degree), linear, or moving-average. Excel will plot the regression curve against your data automatically, and (except for moving average) you can tell it to give you an equation for the curve. To do this:
Right click on one of the data points (it doesn't matter which). Click "add trendline..."
A new window will come up, asking you for the type of regression. Choose the type of regression you want to use.
Click on the "options" tab, and check the "Display Equation on Chart" box (and, if you want, the "Display R-squared value on Chart" box). Click OK.
If you chose a "linear" regression with the sample data above, the equation and
{\displaystyle R^{2}}
value appear on the graph as
{\displaystyle y=177.87x+391.28,R^{2}=0.9843}
. Note Excel displays
{\displaystyle R^{2}}
rather than R (so that we don't need to worry about negative vs. positive values); if you want R just take the square root, which is 0.9921 as we calculated in the section on linear regressions.
Further resources for SpreadsheetsEdit
Excel and other spreadsheets can do far more than what is described here. For additional information, see Microsoft Office, w:Excel, or the help files for the program you are using.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Introduction_to_Chemical_Engineering_Processes/Excel&oldid=3325778"
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Proposition 69.3.10 (04AK)—The Stacks project
Chapter 69: Limits of Algebraic Spaces
Proposition 69.3.10. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:
The morphism $f$ is a morphism of algebraic spaces which is locally of finite presentation, see Morphisms of Spaces, Definition 66.28.1.
The morphism $f : X \to Y$ is limit preserving as a transformation of functors, see Definition 69.3.1.
Proof. Assume (1). Let $T$ be a scheme and let $y \in Y(T)$. We have to show that $T \times _ Y X$ is limit preserving over $T$ in the sense of Definition 69.3.1. Hence we are reduced to proving that if $X$ is an algebraic space which is locally of finite presentation over $S$ as an algebraic space, then it is limit preserving as a functor $X : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. To see this choose a presentation $X = U/R$, see Spaces, Definition 64.9.3. It follows from Morphisms of Spaces, Definition 66.28.1 that both $U$ and $R$ are schemes which are locally of finite presentation over $S$. Hence by Limits, Proposition 32.6.1 we have
\[ U(T) = \mathop{\mathrm{colim}}\nolimits U(T_ i), \quad R(T) = \mathop{\mathrm{colim}}\nolimits R(T_ i) \]
whenever $T = \mathop{\mathrm{lim}}\nolimits _ i T_ i$ in $(\mathit{Sch}/S)_{fppf}$. It follows that the presheaf
\[ (\mathit{Sch}/S)_{fppf}^{opp} \longrightarrow \textit{Sets}, \quad W \longmapsto U(W)/R(W) \]
is limit preserving. Hence by Lemma 69.3.7 its sheafification $X = U/R$ is limit preserving too.
Assume (2). Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Next, choose a scheme $U$ and a surjective étale morphism $U \to V \times _ Y X$. By Lemma 69.3.5 the transformation of functors $V \times _ Y X \to V$ is limit preserving. By Morphisms of Spaces, Lemma 66.39.8 the morphism of algebraic spaces $U \to V \times _ Y X$ is locally of finite presentation, hence limit preserving as a transformation of functors by the first part of the proof. By Lemma 69.3.3 the composition $U \to V \times _ Y X \to V$ is limit preserving as a transformation of functors. Hence the morphism of schemes $U \to V$ is locally of finite presentation by Limits, Proposition 32.6.1 (modulo a set theoretic remark, see last paragraph of the proof). This means, by definition, that (1) holds.
Set theoretic remark. Let $U \to V$ be a morphism of $(\mathit{Sch}/S)_{fppf}$. In the statement of Limits, Proposition 32.6.1 we characterize $U \to V$ as being locally of finite presentation if for all directed inverse systems $(T_ i, f_{ii'})$ of affine schemes over $V$ we have $U(T) = \mathop{\mathrm{colim}}\nolimits V(T_ i)$, but in the current setting we may only consider affine schemes $T_ i$ over $V$ which are (isomorphic to) an object of $(\mathit{Sch}/S)_{fppf}$. So we have to make sure that there are enough affines in $(\mathit{Sch}/S)_{fppf}$ to make the proof work. Inspecting the proof of (2) $\Rightarrow $ (1) of Limits, Proposition 32.6.1 we see that the question reduces to the case that $U$ and $V$ are affine. Say $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(B)$. By construction of $(\mathit{Sch}/S)_{fppf}$ the spectrum of any ring of cardinality $\leq |B|$ is isomorphic to an object of $(\mathit{Sch}/S)_{fppf}$. Hence it suffices to observe that in the "only if" part of the proof of Algebra, Lemma 10.127.3 only $A$-algebras of cardinality $\leq |B|$ are used. $\square$
Comment #2298 by Eric Ahlqvist on November 07, 2016 at 13:26
Hi! I think you should switch
X
Y
in the fiber product on the first line of the proof:
T\times_XY
T\times_YX
In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04AK. Beware of the difference between the letter 'O' and the digit '0'.
The tag you filled in for the captcha is wrong. You need to write 04AK, in case you are confused.
|
PicoCTF transformation | Ari
Breaking into things isn't always as easy as it sounds. In this post, we're going to take a peek at an "easy-ish" reverse engineering challenge that requires some knowledge of bit manipulation.
Why would you ever need to know about bit manipulation? If you wanted to encrypt or encode a message to someone, bit manipulation is a great way to start off.
The challenge we're working on today is located at https://play.picoctf.org/practice/challenge/104?page=1. The description of the challenge reads as follows:
I wonder what this really is... enc
Hm... that's a little confusing, right? Well... let's take a look inside the enc file that's included in challenge. It seems to be a bunch of Chinese characters.
λ cat enc
灩捯䍔䙻ㄶ形楴獟楮獴㌴摟潦弸弰㑣〷㘰摽%
What about the python code included?
To understand what's happening here, let's break it up into smaller pieces and rewrite it so it's not a crazy list comprehension (that can be hard to read):
code1 = ord(flag[i]) << 8
char1 = chr(code1)
code2 = code1 + ord(flag[i + 1])
char2 = chr(code)
s += char1 + char2
The original version feels complex to read, so breaking it up into English, the enc file is "encrypted-ish" as the following (in English):
For every 2 characters, shift the first unicode character to the left by 8 bits and append that to the string and then take the 2nd unicode number and add it to the first unicode number and append that character to a string. Phew... that's less easy to understand, eh?
What's going on with that << 8 thing anyway?
Short introduction to bit manipulation
Although it sounds scary, bit manipulation is kind of how computers do most things (kinda), but it's an important component of how computers do most of the things they do.
Starting off with the decimal system, like numbers 1, 10, 134, etc. is the decimal system. The decimal system is a base-10 numbering system.
There are other numbering systems as well. The one we're going to look at for this post is the base-2 system: binary
Let's look at the basics of binary. Using one byte (four bits -- or just four 1s or 0s), let's look at some values:
The number 0 is just four 0s in a row: 0000. The number 1 is three 0s and a 1: 0001. Okay, so what's two?
Two more examples, what's 4 and 5?
Looking at the decimal system, our usual base-10 system, we can see the same pattern where the placement of the number adds up to the total. Each placement in decimal is an order of 10. So
10^0
is in the 1s spot,
10^1
is in the 10s place,
10^2
is in the 100s place, and so on and so forth:
242 => 200 + 40 + 2
In binary, the placement of each number is an order of the number 2 (base-2). Each placement in the binary system is set at it's 2^0 characteristic.
0001 => 1 + 0 + 0 + 0
The first number (in the same direction as binary) is
2^0
=> 1. The second is
2^1
=> 2, the third is
2^2
=> 4. With each successive number, we're just adding to the total.
For instance, how do we express 9?
8 + 1, so we need a 1 in the 8 spot and a 1 in the 1 place: 1001.
Okay, but how about that challenge?
Now that we know what's happening with the underlying numbers, we need to know how computers express the ascii digits (English characters) and unicode digits (way cooler characters, like cyrillic and mandarin characters... even emojis).
ASCII characters is a way for computers to store characters as numbers (and it's an old standard)
Unicode is a way of storing numbers that express all sorts of different characters
When you see a website that has emojis, it's supporting unicode
This matters because in our challenge, we're taking characters in the Mandarine characters and transforming them from unicode to ascii. In python, we have two different operators that will help us (and are in the hint above):
ord() takes a unicode character and returns it's unicode value
chr() takes a number and returns the ascii value
In python, converting from ASCII character strings into numbers we can use the ord() function:
python -c "print(ord('A'))" # 65
To convert from unicode characters to character values, we have to do a little more work, but we'll use a combination of the ord() function and the chr() function:
λ python -c "print(ord('灩'))" # 28777
λ python -c "print(chr('灩'))" # ERROR TypeError: an integer is required (got type str)
We'll need to somehow convert this unicode character into an ascii-readable character. Using our bit-manipulation technique from above, let's shift the unicode values by 8 over to the left (shrinking the value, because unicode characters are MUCH larger than ascii characters) and then call chr() on the resulting value:
λ python -c "print(ord('灩') >> 8)" # 112
λ python -c "print(chr(ord('灩') >> 8))" # p
Sweet. Now we can take each character in our string and reverse-engineer the original encoding (that's the code we translated above).
Using python, we'll read the file into a variable we'll call flag:
with open('enc', 'r') as fp:
flag = fp.read()
Now for each character, we'll convert the first value into an ascii representation:
f = ord(flag[i]) >> 8
s += chr(f)
Hm... when we run this, the output is too short and doesn't follow the flag pattern starting with picoCTF{}.
λ ./blog.py
pcCF1_isis3do__406d
Ah we forgot that in the original encoding, the second character is added to the ordinal value to create the character encoding. We need to reverse this process and subtract the second character.
f = ord(flag[i]) >> 8 # first character ascii value
t = ((ord(flag[i])) - ((ord(flag[i]) >> 8))) # second character ascii value
s += chr(t)
On the line where we're calculating the second value, we're going to get an error. The error we'll get occurs because we're still try to convert a weird (or negative) unicode value. Instead, we need to subtract an ascii value. We'll just need to shift our character back by 8 (unicode to ascii).
t = ((ord(flag[i])) - ((ord(flag[i]) >> 8) << 8)) # secondcharacter ascii value
Let's run our reverse engineering script:
picoCTF{16_bits_inst34d_of_8_XXXXXXXX} # scrubbed answer
Awesome. We got the flag. That's 20 more points.
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